The $\lambda$-additive measure in a new light: the $Q_{\nu }$ measure and its connections with belief, probability, plausibility, rough sets, multi-attribute utility functions and fuzzy operators

Abstract

The aim of this paper is twofold. On the one hand, the $\lambda$-additive measure (Sugeno $\lambda$-measure) is revisited, and a state-of-the-art summary of its most important properties is provided. On the other hand, the so-called $\nu$-additive measure as an alternatively parameterized $\lambda$-additive measure is introduced. Here, the advantages of the $\nu$-additive measure are discussed, and it is demonstrated that these two measures are closely related to various areas of science. The motivation for introducing the $\nu$-additive measure lies in the fact that its parameter $\nu \in (0,1)$ has an important semantic meaning as it is the fix point of the complement operation. Here, by utilizing the $\nu$-additive measure, some well-known results concerning the $\lambda$-additive measure are put into a new light and rephrased in more advantageous forms. It is discussed here how the $\nu$-additive measure is connected with the belief-, probability- and plausibility measures. Next, it is also shown that two $\nu$-additive measures, with the parameters $\nu _1$ and $\nu _2$, are a dual pair of belief- and plausibility measures if and only if $\nu _1+\nu _2 = 1$. Furthermore, it is demonstrated how a $\nu$-additive measure (or a $\lambda$-additive measure) can be transformed to a probability measure and vice versa. Lastly, it is discussed here how the $\nu$-additive measures are connected with rough sets, multi-attribute utility functions and certain operators of fuzzy logic.

1 Introduction

It is an acknowledged fact that the $\lambda$-additive measure (Sugeno $\lambda$-measure) (Sugeno 1974) is one of the most widely applied monotone measures (fuzzy measure). The usefulness, versatility and applicability of $\lambda$-additive measures have inspired numerous theoretical and practical researches since Sugeno’s original results were published in 1974 (see, e.g., Magadum and Bapat 2018; Mohamed and Xiao 2003; Chiţescu 2015; Chen et al. 2016; Singh 2018).

The aim of the present study is twofold. On the one hand, we will revisit the $\lambda$-additive measure and give a state-of-the-art summary of its most important properties. On the other hand, we will introduce the so-called $\nu$-additive measure as an alternatively parameterized $\lambda$-additive measure, demonstrate the advantages of the $\nu$-additive measure and point out that these two measures are closely related to various areas of science. The motivation for introducing the $\nu$-additive measure lies in the fact that its parameter $\nu \in (0,1)$ has an important semantic meaning. Namely, $\nu$ is the fix point of the complement operation; that is, if the $\nu$ additive measure of a set has the value $\nu$, then the $\nu$-additive measure of its complement set has the value $\nu$ as well. It should be added that by utilizing the $\nu$-additive measure, some well-known results concerning the $\lambda$-additive measure can be put into a new light and rephrased in more advantageous forms. Here, we will discuss how the $\nu$-additive measure is connected with the belief-, probability- and plausibility measures (see, e.g., Wang and Klir 2013; Höhle 1987; Dubois and Prade 1980; Spohn 2012; Feng et al. 2014). Also, we will demonstrate that a $\nu$-additive measure is a

• (1) belief measure if and only if $0<\nu \le 1/2$

• (2) probability measure if and only if $\nu =1/2$

• (3) plausibility measure if and only if $1/2 \le \nu <1$.

Next, we will show that two $\nu$-additive measures, with the parameters $\nu _1$ and $\nu _2$, are a dual pair of belief- and plausibility measures if and only if $\nu _1+\nu _2 = 1$. Furthermore, we will discuss how a $\nu$-additive measure (or a $\lambda$-additive measure) can be transformed to a probability measure and vice versa. Moreover, we will also discuss how the $\nu$-additive measures are connected with rough sets (see, e.g., Dubois and Prade 1990; Yao and Lingras 1998; Wu et al. 2002; Polkowski 2013), with multi-attribute utility functions (see, e.g., Sarin 2013; Greco et al. 2016; Keeney and Raiffa 1993), and with certain operators of continuous-valued logic (see, e.g., Dombi 1982, 2008).

The rest of this paper is structured as follows: In Sect. 2, we give an overview of the monotone (fuzzy) measures including the belief-, probability- and plausibility measures. In Sect. 3, the $\nu$-additive measure is introduced and its key properties are discussed. In Sect. 4, we demonstrate how the $\nu$-additive measure is related to the belief-, probability- and plausibility measures, and in Sect. 5, we show how a $\nu$-additive measure can be transformed to a probability measure and vice versa. Section 6 reveals some areas of science which the $\nu$-additive ($\lambda$-additive) measures are connected with. Lastly, in Sect. 7, we give a short summary of our findings and highlight our future research plans including the possible application of $\nu$-additive measure in network science.

In this study, we will use the common notations $\cap$ and $\cup$ for the intersection and union operations over sets, respectively. Also, will use the notation $\overline{A}$ for the complement of set A.

2 Monotone measures

Now, we will introduce the monotone measures and give a short overview of them that covers the probability-, belief- and plausibility measures.

Definition 1

Let $\varSigma$ be a $\sigma$-algebra on the set X. Then, the function $g: \varSigma \rightarrow [0,1]$ is a monotone measure on the measurable space $(X, \varSigma )$ iff g satisfies the following requirements:

• (1) $g(\emptyset ) = 0$, $g(X) = 1$

• (2) if $B \subseteq A$, then $g(B) \le g(A)$ for any $A, B \in \varSigma$ (monotonicity)

• (3) if $\forall i \in \mathbb {N}, A_i \in \varSigma$ and $(A_i)$ is monotonic $(A_1 \subseteq A_2 \subseteq \cdots \subseteq A_n \subseteq \cdots$ or $A_1 \supseteq A_2 \supseteq \cdots \supseteq A_n \cdots )$, then

  $$\begin{aligned} \lim \limits _{i \rightarrow \infty } g(A_i) = g\left( \lim \limits _{i \rightarrow \infty } A_{i} \right) \text {(continuity).} \end{aligned}$$

If X is a finite set, then the continuity requirement in Definition 1 can be disregarded and the monotone measure is defined as follows.

Definition 2

The function $g: \mathscr {P}(X) \rightarrow [0,1]$ is a monotone measure on the finite set X iff g satisfies the following requirements:

• (1) $g(\emptyset ) = 0$, $g(X) = 1$

• (2) if $B \subseteq A$, then $g(B) \le g(A)$ for any $A, B \in \mathscr {P}(X)$ (monotonicity).

Note that the monotone measures given by Definitions 1 and 2 are known as fuzzy measures, which were originally defined by Choquet (1954) and Sugeno (1974).

2.1 Some examples of monotone measures

2.1.1 Dirac measure

Definition 3

The function $\delta _{x_{0}}: \mathscr {P}(X) \rightarrow [0,1]$ is a Dirac measure on the set X, iff $\forall A \in \mathscr {P}(X)$:

$$\begin{aligned} \delta _{x_{0}}(A) = {\left\{ \begin{array}{ll} 1, &{} \text {if }\, x_{0} \in A\\ 0, &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

2.1.2 Probability measure

Definition 4

Let $\varSigma$ be a $\sigma$-algebra over the set X. Then, the function $Pr: \varSigma \rightarrow [0,1]$ is a probability measure on the space $(X, \varSigma )$ iff Pr satisfies the following requirements:

• (1) $\forall A \in \varSigma : Pr(A) \ge 0$

• (2) $Pr(X) = 1$

• (3) $\forall A_{1}, A_{2}, \ldots \in \varSigma$, if $A_i \cap A_j = \emptyset , \forall i\ne j$, then

  $$\begin{aligned} Pr\left( \bigcup \limits _{i=1}^{\infty } A_i \right) =\sum \limits _{i=1}^{\infty }Pr(A_i). \end{aligned}$$

Remark 1

If X is a finite set, then requirement (3) in Definition 4 can be reduced to the following requirement: for any disjoint $A, B \in \mathscr {P}(X)$, $Pr(A \cup B) = Pr(A) + Pr(B)$.

2.1.3 Belief measure and plausibility measure

Definition 5

The function $Bl: \mathscr {P}(X) \rightarrow [0,1]$ is a belief measure on the finite set X, iff Bl satisfies the following requirements:

• (1) $Bl(\emptyset ) = 0, Bl(X) = 1$

• (2) for any $A_1, A_2, \ldots , A_n \in \mathscr {P}(X)$,

  $$\begin{aligned} \begin{aligned}&Bl( A_1 \cup A_2 \cup \cdots \cup A_n) \\&\quad \ge \sum \limits _{k=1}^n \sum \limits _{\begin{array}{c} 1 \le i_{1}< i_{2} \cdots \\ \cdots < i_{k} \le n \end{array}} (-1)^{k-1} Bl \left( A_{i_{1}} \cap A_{i_{1}} \cap \cdots \cap A_{i_{k}} \right) . \end{aligned} \end{aligned}$$

  (1)

Here, Bl(A) is interpreted as a grade of belief in that a given element of X belongs to A.

Lemma 1

If Bl is a belief measure on the finite set X, then for any $A \in \mathscr {P}(X)$,

$$\begin{aligned} Bl(A)+ Bl(\overline{A}) \le 1. \end{aligned}$$

Proof

Noting Definition 5, we have

$$\begin{aligned} 1 = Bl(A \cup \overline{A}) \ge Bl(A) + Bl(\overline{A}) - Bl(A \cap \overline{A}) = Bl(A) + Bl(\overline{A}). \end{aligned}$$

$\square$

The inequality $Bl(A) + Bl(\overline{A}) \le 1$ means that a lack of belief in $x \in A$ does not imply a strong belief in $x \in \overline{A}$. In particular, total ignorance is modeled by the belief function $Bl_i$ such that $Bl_i(A) = 0$ if $A\ne X$ and $Bl_i(A) = 1$ if $A = X$.

The following proposition is about the monotonicity of belief measures.

Proposition 1

If X is a finite set, Bl is a belief measure on X, $A,B \in \mathscr {P}(X)$ and $B \subseteq A$, then $Bl(B) \le Bl(A)$.

Proof

Let $B \subseteq A$. Hence, there exists $C \in \mathscr {P}(X)$ such that $A = B\cup C$ and $B \cap C = \emptyset$. Now, by utilizing the definition of the belief measure and the fact that $B \cap C = \emptyset$, we get

$$\begin{aligned} Bl(A) = Bl(B \cup C) \ge Bl(B) + Bl(C) \ge Bl(B). \end{aligned}$$

$\square$

Corollary 1

The belief measure given by Definition 5 is a monotone measure.

Proof

Let Bl be a belief measure. It follows from Definition 5 that Bl satisfies criterion (1) for a monotone measure given in Definition 2. Moreover, the monotonicity of Bl is proven in Proposition 1; that is, Bl also satisfies criterion (2) in Definition 2. $\square$

Definition 6

The function $Pl: \mathscr {P}(X) \rightarrow [0,1]$ is a plausibility measure on the finite set X, iff Pl satisfies the following requirements:

• (1) $Pl(\emptyset ) = 0, Pl(X) = 1$

• (2) for any $A_1, A_2, \ldots , A_n \in \mathscr {P}(X)$,

  $$\begin{aligned} \begin{aligned}&Pl( A_1 \cap A_2 \cap \cdots \cap A_n) \\&\quad \le \sum \limits _{k=1}^n \sum \limits _{\begin{array}{c} 1 \le i_{1}< i_{2} \cdots \\ \cdots < i_{k} \le n \end{array}} (-1)^{k-1} Pl \left( A_{i_{1}} \cup A_{i_{2}} \cdots \cup A_{i_{k}} \right) . \end{aligned} \end{aligned}$$

  (2)

Here, Pl(A) is interpreted as the plausibility of A.

Lemma 2

If Pl is a plausibility measure on the finite set X, then for any $A \in \mathscr {P}(X)$,

$$\begin{aligned} Pl(A)+ Pl(\overline{A}) \ge 1. \end{aligned}$$

Proof

Noting Definition 6, we have

$$\begin{aligned} \begin{aligned} 0&= Pl(A \cap \overline{A}) \le Pl(A) + Pl(\overline{A}) - Pl(A \cup \overline{A}) \\&= Pl(A) + Pl(\overline{A}) - 1, \end{aligned} \end{aligned}$$

from which $Pl(A)+ Pl(\overline{A}) \ge 1$ follows. $\square$

This result can be interpreted so that the plausibility of $x \in A$ does not imply a strong plausibility of $x \in \overline{A}$.

The following proposition is about the monotonicity of plausibility measures.

Proposition 2

If X is a finite set, Pl is a plausibility measure on X, $A,B \in \mathscr {P}(X)$ and $B \subseteq A$, then $Pl(B) \le Pl(A)$.

Proof

Let $B \subseteq A$. Let $C \in \mathscr {P}(X)$ such that $A \cap C = B$ and $A \cup C = X$. Now, by utilizing the definition of plausibility measure, and the facts that $A \cup C = X$ and $Pl(C) \le 1$, we get

$$\begin{aligned} \begin{aligned} Pl(B)&= Pl(A \cap C) \le Pl(A)+Pl(C) - Pl(A \cup C) \\&= Pl(A)+Pl(C) -1 \le Pl(A). \end{aligned} \end{aligned}$$

$\square$

Corollary 2

The plausibility measure given by Definition 6 is a monotone measure.

Proof

Let Pl be a plausibility measure. It follows from Definition 6 that Pl satisfies criterion (1) for a monotone measure given in Definition 2. Next, the monotonicity of Pl is proven in Proposition 2; that is, Pl also satisfies criterion (2) in Definition 2. $\square$

The plausibility of a subset A of the finite set X was defined by Shafer (1976) as

$$\begin{aligned} Pl(A) = 1 - Bl( \overline{A}), \end{aligned}$$

where Bl is a belief function. The following proposition states an interesting connection between the belief measure and the plausibility measure.

Proposition 3

Let X be a finite set and let $\mu _1, \mu _2: \mathscr {P}(X) \rightarrow [0,1]$ be two monotone measures on X such that

$$\begin{aligned} \mu _2(A) = 1 - \mu _1( \overline{A}) \end{aligned}$$

(3)

holds for any $A \in \mathscr {P}(X)$. Then, either (1) $\mu _1$ is a belief measure on X if and only if $\mu _2$ is a plausibility measure on X, or (2) $\mu _1$ is a plausibility measure on X if and only if $\mu _2$ is a belief measure on X.

Proof

We will prove case (1), and the proof of case (2) is similar. Firstly, we will show that if $\mu _1$ is a belief measure on X and $\mu _2(A)$ is given as $\mu _2(A) = 1-\mu _1(\overline{A})$ for any $A \in \mathscr {P}(X)$, then $\mu _2$ is a plausibility measure on X. Let $\mu _1$ be a belief measure on X and $\mu _2(A) = 1-\mu _1(\overline{A})$ for any $A \in \mathscr {P}(X)$. Then, $\mu _2(\emptyset )=0$ and $\mu _2(X)=1$ trivially follow from the fact that $\mu _1$ is a belief measure and $\mu _2(A) = 1-\mu _1(\overline{A})$. That is, function $\mu _2$ satisfies requirement (1) for a plausibility measure given in Definition 6. Furthermore, since function $\mu _1$ is a belief measure, the inequality

$$\begin{aligned} \begin{aligned}&\mu _1( A_1 \cup A_2 \cup \cdots \cup A_n) \\&\quad \ge \sum \limits _{k=1}^n \sum \limits _{\begin{array}{c} 1 \le i_{1}< i_{2} \cdots < i_{k} \le n \end{array}} (-1)^{k-1} \mu _1 \left( A_{i_{1}} \cap A_{i_{1}} \cap \cdots \cap A_{i_{k}} \right) . \end{aligned} \end{aligned}$$

(4)

holds for any $A_1, A_2, \ldots , A_n \in \mathscr {P}(X)$. From the condition $\mu _2(A) = 1-\mu _1(\overline{A})$, we also have that $\mu _1(A) = 1-\mu _2(\overline{A})$. Next, applying the inequality in Eq. (4) to the complement sets $\overline{A_1}, \overline{A_2}, \ldots , \overline{A_n} \in \mathscr {P}(X)$ and utilizing the fact that $\mu _1(A) = 1-\mu _2(\overline{A})$, we get

$$\begin{aligned} \begin{aligned}&1-\mu _2(\overline{\overline{A_1} \cup \overline{A_2} \cup \cdots \cup \overline{A_n}}) \\&\quad \ge 1-\mu _2(\overline{\overline{A_1}}) + 1-\mu _2(\overline{\overline{A_2}}) + \cdots + 1-\mu _2(\overline{\overline{A_n}}) \\&\qquad - (1 - \mu _2(\overline{\overline{A_1} \cap \overline{A_2}})) - \cdots -(1 - \mu _2(\overline{\overline{A_{n-1}} \cap \overline{A_n}})) + \cdots \\&\qquad + (-1)^{n+1} \left( 1- \mu _2(\overline{\overline{A_1} \cap \overline{A_2} \cap \cdots \cap \overline{A_n}}) \right) \\&\quad = -\mu _2(\overline{\overline{A_1}}) -\mu _2(\overline{\overline{A_2}}) - \cdots -\mu _2(\overline{\overline{A_n}}) \\&\qquad + \mu _2(\overline{\overline{A_1} \cap \overline{A_2}}) + \cdots + \mu _2(\overline{\overline{A_{n-1}} \cap \overline{A_n}}) + \cdots \\&\qquad + (-1)^{n} \mu _2(\overline{\overline{A_1} \cap \overline{A_2} \cap \cdots \cap \overline{A_n}}) \\&\qquad + \left( {\begin{array}{c}n\\ 1\end{array}}\right) - \left( {\begin{array}{c}n\\ 2\end{array}}\right) + \cdots +\left( {\begin{array}{c}n\\ n\end{array}}\right) (-1)^{n+1}. \end{aligned} \end{aligned}$$

Noting the fact that

$$\begin{aligned} \sum _{k=1}^{n} \left( {\begin{array}{c}n\\ k\end{array}}\right) (-1)^{k+1} = 1, \end{aligned}$$

(5)

the previous inequality can be written as

$$\begin{aligned} \begin{aligned}&1-\mu _2(\overline{\overline{A}_1 \cup \overline{A}_2 \cup \cdots \cup \overline{A}_n}) \\&\quad \ge 1 -\mu _2(\overline{\overline{A_1}}) -\mu _2(\overline{\overline{A_2}}) - \cdots -\mu _2(\overline{\overline{A_n}}) \\&\qquad + \mu _2(\overline{\overline{A_1} \cap \overline{A_2}}) + \cdots + \mu _2(\overline{\overline{A_{n-1}} \cap \overline{A_n}}) + \cdots \\&\qquad + (-1)^{n} \mu _2(\overline{\overline{A_1} \cap \overline{A_2} \cap \cdots \cap \overline{A_n}}). \end{aligned} \end{aligned}$$

Now, applying the De Morgan law to the last inequality, we get

$$\begin{aligned} \begin{aligned}&\mu _2(A_1 \cap A_2 \cap \cdots \cap A_n) \\&\quad \le \sum \limits _{k=1}^n \sum \limits _{\begin{array}{c} 1 \le i_{1}< i_{2} \cdots < i_{k} \le n \end{array}} (-1)^{k-1} \mu _2 \left( A_{i_{1}} \cup A_{i_{1}} \cup \cdots \cup A_{i_{k}} \right) , \end{aligned} \end{aligned}$$

which means that function $\mu _2$ is a plausibility measure.

Secondly, we will demonstrate that if $\mu _2$ is a plausibility measure on X and $\mu _2(A)$ is given as $\mu _2(A) = 1-\mu _1(\overline{A})$ for any $A \in \mathscr {P}(X)$, then $\mu _1$ is a belief measure on X. Let $\mu _2$ be a plausibility measure on X and $\mu _2(A) = 1-\mu _1(\overline{A})$ for any $A \in \mathscr {P}(X)$. These conditions trivially imply that $\mu _1(\emptyset )=0$ and $\mu _1(X)=1$; that is, function $\mu _1$ satisfies requirement (1) for a belief measure given in Definition 5. Next, because function $\mu _2$ is a plausibility measure, the inequality

$$\begin{aligned} \begin{aligned}&\mu _2( A_1 \cap A_2 \cap \cdots \cap A_n) \\&\quad \le \sum \limits _{k=1}^n \sum \limits _{\begin{array}{c} 1 \le i_{1}< i_{2} \cdots < i_{k} \le n \end{array}} (-1)^{k-1} \mu _2 \left( A_{i_{1}} \cup A_{i_{2}} \cdots \cup A_{i_{k}} \right) \end{aligned} \end{aligned}$$

(6)

holds for any $A_1, A_2, \ldots , A_n \in \mathscr {P}(X)$. Then, applying the inequality in Eq. (6) to the complement sets $\overline{A_1}, \overline{A_2}, \ldots , \overline{A_n} \in \mathscr {P}(X)$ and utilizing the condition that $\mu _2(A) = 1-\mu _1(\overline{A})$, we get

$$\begin{aligned} \begin{aligned}&1-\mu _1(\overline{\overline{A_1} \cap \overline{A_2} \cap \cdots \cap \overline{A_n}}) \\&\quad \le 1-\mu _1(\overline{\overline{A_1}}) + 1-\mu _1(\overline{\overline{A_2}}) + \cdots + 1-\mu _1(\overline{\overline{A_n}}) \\&\qquad - (1 - \mu _1(\overline{\overline{A_1} \cup \overline{A_2}})) - \cdots -(1 - \mu _1(\overline{\overline{A_{n-1}} \cup \overline{A_n}})) +\cdots \\&\qquad + (-1)^{n+1} \left( 1- \mu _1(\overline{\overline{A_1} \cup \overline{A_2} \cup \cdots \cup \overline{A_n}}) \right) \\&\quad = -\mu _1(\overline{\overline{A_1}}) -\mu _1(\overline{\overline{A_2}}) - \cdots -\mu _1(\overline{\overline{A_n}}) \\&\qquad + \mu _1(\overline{\overline{A_1} \cup \overline{A_2}}) + \cdots + \mu _1(\overline{\overline{A_{n-1}} \cup \overline{A_n}}) + \cdots \\&\qquad + (-1)^{n} \mu _1(\overline{\overline{A_1} \cup \overline{A_2} \cup \cdots \cup \overline{A_n}}) \\&\qquad + \left( {\begin{array}{c}n\\ 1\end{array}}\right) - \left( {\begin{array}{c}n\\ 2\end{array}}\right) + \cdots +\left( {\begin{array}{c}n\\ n\end{array}}\right) (-1)^{n+1}. \end{aligned} \end{aligned}$$

Again, taking into account Eq. (5), the previous inequality can be written as

$$\begin{aligned} \begin{aligned}&1-\mu _1(\overline{\overline{A}_1 \cap \overline{A}_2 \cap \cdots \cap \overline{A}_n}) \\&\quad \le 1 -\mu _1(\overline{\overline{A_1}}) -\mu _1(\overline{\overline{A_2}}) - \cdots -\mu _1(\overline{\overline{A_n}}) \\&\qquad + \mu _1(\overline{\overline{A_1} \cup \overline{A_2}}) + \cdots + \mu _1(\overline{\overline{A_{n-1}} \cup \overline{A_n}}) + \cdots \\&\qquad + (-1)^{n} \mu _1(\overline{\overline{A_1} \cup \overline{A_2} \cup \cdots \cup \overline{A_n}}). \end{aligned} \end{aligned}$$

Now, applying the De Morgan law to the last inequality, we get

$$\begin{aligned} \begin{aligned}&\mu _1(A_1 \cup A_2 \cup \cdots \cup A_n) \\&\quad \ge \sum \limits _{k=1}^n \sum \limits _{\begin{array}{c} 1 \le i_{1}< i_{2} \cdots < i_{k} \le n \end{array}} (-1)^{k-1} \mu _1 \left( A_{i_{1}} \cap A_{i_{1}} \cap \cdots \cap A_{i_{k}} \right) . \end{aligned} \end{aligned}$$

Hence, $\mu _1$ is a belief measure. $\square$

Later, we will use the concept of dual pair of belief- and plausibility measures.

Definition 7

Let Bl and Pl be a belief measure and a plausibility measure, respectively, on set X. Then Bl and Pl are said to be a dual pair of belief- and plausibility measures iff

$$\begin{aligned} Pl(A) = 1 - Bl( \overline{A}) \end{aligned}$$

holds for any $A \in \mathscr {P}(X)$.

In the Dempster–Shafer theory of evidence, a belief mass is assigned to each element of the power set $\mathscr {P}(X)$, where X is a finite set. The belief mass is given by the so-called basic probability assignment m from $\mathscr {P}(X)$ to [0, 1] that is defined as follows.

Definition 8

The function $m: \mathscr {P}(X) \rightarrow [0,1]$ is a basic probability assignment (mass function) on the finite set X, iff m satisfies the following requirements:

• (1) $m (\emptyset ) = 0$

• (2) $\sum _{A\in \mathscr {P}(X)} m(A) = 1$.

The subsets A of X for which $m(A) > 0$ are called the focal elements of m. Let $x \in A$ and $A \in \mathscr {P}(X)$. Then, the mass m(A) can be interpreted as the probability of knowing $x \in A$ given the available evidence. Utilizing a given basic probability assignment m, the belief Bl(A) for the set A is

$$\begin{aligned} Bl(A) = \sum \limits _{B \vert B \subseteq A} m(B), \end{aligned}$$

and the plausibility Pl(A) is

$$\begin{aligned} Pl(A) = \sum \limits _{B \vert B \cap A \ne \emptyset } m(B). \end{aligned}$$

A basic probability assignment m can be represented by its belief function Bl as

$$\begin{aligned} m(B) = \sum \limits _{A \subseteq B} (-1)^{\vert B \setminus A \vert } Bl(A), \end{aligned}$$

where $B \in \mathscr {P}(X)$. Here, m is the basic probability assignment of the belief measure Bl. Note that plausibility measures and belief functions were introduced by Dempster (1967) under the names upper and lower probabilities, induced by a probability measure by a multi-valued mapping.

Remark 2

The monotonicity of the plausibility measure Pl can also be demonstrated by utilizing the duality $Pl(A) = 1-Bl(\overline{A})$ and the monotonicity of the belief measure Bl. Namely, if $B \subseteq A$, then $\overline{A} \subseteq \overline{B}$ and so

$$\begin{aligned} Bl(\overline{A}) \le Bl(\overline{B}), \end{aligned}$$

from which

$$\begin{aligned} 1-Bl(\overline{A}) \ge 1-Bl(\overline{B}), \end{aligned}$$

which means that

$$\begin{aligned} Pl(A) \ge Pl(B). \end{aligned}$$

3 Introduction to the $Q_{\nu }$ measure

Relaxing the additivity property of the probability measure, the $\lambda$-additive measures were proposed by Sugeno (1974).

Definition 9

The function $Q_{\lambda }: \mathscr {P}(X) \rightarrow [0,1]$ is a $\lambda$-additive measure (Sugeno $\lambda$-measure) on the finite set X, iff $Q_{\lambda }$ satisfies the following requirements:

• (1) $Q_{\lambda }(X) = 1$

• (2) for any $A, B \in \mathscr {P}(X)$ and $A \cap B=\emptyset$,

  $$\begin{aligned} Q_{\lambda }(A \cup B)=Q_{\lambda }(A)+Q_{\lambda }(B)+\lambda Q_{\lambda }(A)Q_{\lambda }(B), \end{aligned}$$

  (7)

  where $\lambda \in (-1, \infty )$.

Note that if X is an infinite set, then the continuity of function $Q_{\lambda }$ is also required. Here, we will show that the $\lambda$-additive measures are monotone measures as well.

Proposition 4

Every $\lambda$-additive measure is a monotone measure.

Proof

Let $Q_{\lambda }$ be a $\lambda$-additive measure on the set X. Then $Q_{\lambda }(X)=1$ holds by definition. Next, by utilizing Eq. (7), we get $Q_{\lambda }(X)=Q_{\lambda }(X \cup \emptyset ) =Q_{\lambda }(X)+Q_{\lambda }(\emptyset )(1+ \lambda Q_{\lambda }(X))$, which implies that $Q_{\lambda }(\emptyset )=0$. Thus, $Q_{\lambda }$ satisfies criterion (1) of a monotone measure given in Definition 2.

Next, let $A,B \in \mathscr {P}(X)$ and let $B \subseteq A$. Then there exists a $C \in \mathscr {P}(X)$ such that $A = B \cup C$ and $B \cap C=\emptyset$. Now, by utilizing Eq. (7) and the fact that $\lambda > -1$, we get

$$\begin{aligned} \begin{aligned} Q_{\lambda }(A)&= Q_{\lambda }(B \cup C) \\&= Q_{\lambda }(B)+ Q_{\lambda }(C)(1+\lambda Q_{\lambda }(B)) \ge Q_{\lambda }(B). \end{aligned} \end{aligned}$$

It means that $Q_{\lambda }$ also satisfies the monotonicity criterion of a monotone measure. $\square$

Remark 3

The requirement $\lambda \ge -1$ instead of the requirement $\lambda >-1$ would be sufficient to ensure the monotonicity of $Q_{\lambda }$ (see Proposition 4). The requirement $\lambda \ge -1$ also ensures that for any $A, B \in \mathscr {P}(X)$ and $A \cap B=\emptyset$ the $Q_{\lambda }(A \cup B)$ quantity is nonnegative. Namely, since $Q_{\lambda }(A), Q_{\lambda }(B) \in [0,1]$, the inequality

$$\begin{aligned} Q_{\lambda }(A) Q_{\lambda }(B) \le \sqrt{Q_{\lambda }(A) Q_{\lambda }(B)} \le Q_{\lambda }(A) + Q_{\lambda }(B) \end{aligned}$$

holds, and so if $\lambda \ge -1$, then

$$\begin{aligned} \begin{aligned} 0&\le (1+\lambda ) Q_{\lambda }(A) Q_{\lambda }(B) \\&= Q_{\lambda }(A) Q_{\lambda }(B) + \lambda Q_{\lambda }(A) Q_{\lambda }(B) \\&\le Q_{\lambda }(A) + Q_{\lambda }(B) + \lambda Q_{\lambda }(A) Q_{\lambda }(B) = Q_{\lambda }(A \cup B). \end{aligned} \end{aligned}$$

However, the requirement $\lambda >-1$ is given in the definition of $\lambda$-additive measures. Later, we will see that certain properties of $\lambda$-additive measures hold only if $\lambda >-1$.

3.1 The $\lambda$-additive complement and the Dombi form of negation

Proposition 5

If X is a finite set and $Q_{\lambda }$ is a $\lambda$-additive measure on X, then for any $A \in \mathscr {P}(X)$ the $Q_{\lambda }$ measure of the complement set $\overline{A} = X \setminus A$ is

$$\begin{aligned} Q_{\lambda }(\overline{A}) = \frac{1-Q_{\lambda }(A)}{1+\lambda Q_{\lambda }(A)}. \end{aligned}$$

(8)

Proof

Since $A \cap \overline{A}=\emptyset$, we can write

$$\begin{aligned} \begin{aligned} 1&=Q_{\lambda }(X)=Q_{\lambda } (A \cup \overline{A}) \\&= Q_{\lambda }(A)+Q_{\lambda }(\overline{A})+\lambda Q_{\lambda }(A) Q_{\lambda }(\overline{A}) \\&= Q_{\lambda }(A) + Q_{\lambda }(\overline{A})(1+\lambda Q_{\lambda }(A)), \end{aligned} \end{aligned}$$

from which we get

$$\begin{aligned} Q_{\lambda }(\overline{A}) = \frac{1-Q_{\lambda }(A)}{1+\lambda Q_{\lambda }(A)}. \end{aligned}$$

$\square$

Remark 4

For any $A \in \mathscr {P}(X)$, we have

$$\begin{aligned} \begin{aligned}&Q_{\lambda } (A) +Q_{\lambda } (\overline{A})=Q_{\lambda } (A) + \frac{1-Q_{\lambda } (A)}{1+\lambda Q_{\lambda } (A)} \\&\quad = \frac{1+\lambda Q^2_{\lambda } (A)}{1+\lambda Q_{\lambda } (A)}= 1 - \lambda Q_{\lambda } (A)Q_{\lambda } (\overline{A}). \end{aligned} \end{aligned}$$

(9)

It can be seen from Eq. (9) that

$$\begin{aligned} \begin{array}{lll} 0< Q_{\lambda } (A) +Q_{\lambda } (\overline{A}) \le 1 &{} \hbox { if } &{}\quad \lambda \in (0, \infty ) \\ Q_{\lambda } (A) +Q_{\lambda } (\overline{A}) = 1 &{} \hbox { if } &{}\quad \lambda =0 \\ 1 \le Q_{\lambda } (A) +Q_{\lambda } (\overline{A}) < 2 &{}\hbox { if }&{}\quad \lambda \in (-1, 0).\\ \end{array} \end{aligned}$$

We have shown in Proposition 5 that if X is a finite set and $Q_{\lambda }$ is a $\lambda$-additive measure on X, then for any $A \in \mathscr {P}(X)$ the $Q_{\lambda }$ measure of the complement set $\overline{A} = X \setminus A$ is

$$\begin{aligned} Q_{\lambda }(\overline{A}) = \frac{1-Q_{\lambda }(A)}{1+\lambda Q_{\lambda }(A)}. \end{aligned}$$

(10)

Now, let us assume that $0 \le Q(A) <1$. Then, Eq. (10) can be written as

$$\begin{aligned} Q_{\lambda }(\overline{A}) = \frac{1-Q_{\lambda }(A)}{1+\lambda Q_{\lambda }(A)} = \frac{1}{1+(1+\lambda ) \frac{Q_{\lambda }(A)}{1-Q_{\lambda }(A)}}. \end{aligned}$$

(11)

In continuous-valued logic, the Dombi form of negation with the neutral value $\nu \in (0,1)$ is given by the operator $n_{\nu }: [0,1] \rightarrow [0,1]$ as follows:

$$\begin{aligned} n_{\nu }(x) = {\left\{ \begin{array}{ll} \frac{1}{1 + \left( \frac{1 - \nu }{\nu } \right) ^{2} \frac{x}{1 - x}} &{} \text { if }\quad x \in [0,1) \\ 0 &{} \text { if }\quad x=1, \end{array}\right. } \end{aligned}$$

(12)

where $x \in [0,1]$ is a continuous-valued logic variable (Dombi 2008). Note that the Dombi form of negation is the unique Sugeno’s negation (Sugeno 1993) with the fix point $\nu \in (0,1)$. Also, for $Q_{\lambda }(A) \in [0,1)$, the formula of $\lambda$-additive measure of $Q_{\lambda }(\overline{A})$ in Eq. (11) is the same as the formula of the Dombi form of negation in Eq. (12) with $x=Q_{\lambda }(A)$ and

$$\begin{aligned} \left( \frac{1 - \nu }{\nu } \right) ^{2} = 1+\lambda . \end{aligned}$$

Based on the definition of $\lambda$-additive measures, $\lambda >-1$, and since

$$\begin{aligned} \lambda = \left( \frac{1 - \nu }{\nu } \right) ^{2} -1 \end{aligned}$$

is a bijection between (0, 1) and $(-1,\infty )$, the $\lambda$-additive measure of the complement set $\overline{A}$ can be alternatively redefined as

$$\begin{aligned} Q_{\lambda }(\overline{A}) = {\left\{ \begin{array}{ll} \frac{1}{1 + \left( \frac{1 - \nu }{\nu } \right) ^{2} \frac{Q_{\lambda }(A)}{1 - Q_{\lambda }(A)}} &{} \text { if }\quad Q_{\lambda }(A) \in [0,1) \\ 0 &{} \text { if }\quad Q_{\lambda }(A)=1, \end{array}\right. } \end{aligned}$$

(13)

where $\left( \frac{1 - \nu }{\nu } \right) ^{2} = 1+\lambda$, $\nu \in (0,1)$.

Following this line of thinking, here, we will introduce the $\nu$-additive measure and state some of its properties.

Definition 10

The function $Q_{\nu }: \mathscr {P}(X) \rightarrow [0,1]$ is a $\nu$-additive measure on the finite set X, iff $Q_{\nu }$ satisfies the following requirements:

• (1) $Q_{\nu }(X) = 1$

• (2) for any $A, B \in \mathscr {P}(X)$ and $A \cap B=\emptyset$,

  $$\begin{aligned} \begin{aligned}&Q_{\nu }(A \cup B)=Q_{\nu }(A)+Q_{\nu }(B) \\&\quad + \left( \left( \frac{1 - \nu }{\nu } \right) ^{2}-1 \right) Q_{\nu }(A)Q_{\nu }(B), \end{aligned} \end{aligned}$$

  (14)

  where $\nu \in (0, 1)$.

Note that if X is an infinite set, then the continuity of function $Q_{\nu }$ is also required. Here, we state a key proposition that we will frequently utilize later on.

Proposition 6

Let X be a finite set, and let $Q_{\lambda }$ and $Q_{\nu }$ be a $\lambda$-additive and a $\nu$-additive measure on X, respectively. Then,

$$\begin{aligned} Q_{\lambda }(A) = Q_{\nu }(A) \end{aligned}$$

(15)

for any $A \in \mathscr {P}(X)$, if and only if

$$\begin{aligned} \lambda = \left( \frac{1 - \nu }{\nu } \right) ^{2} -1, \end{aligned}$$

(16)

where $\lambda >-1$, $\nu \in (0,1)$.

Proof

This proposition immediately follows from the definitions of the $\lambda$-additive measure and $\nu$-additive measure. $\square$

If $Q_{\nu }$ is a $\nu$-additive measure on the finite set X, then by utilizing Eq. (13), the $Q_{\nu }$ measure of the complement set $\overline{A}$ is

$$\begin{aligned} Q_{\nu }(\overline{A}) = {\left\{ \begin{array}{ll} \frac{1}{1 + \left( \frac{1 - \nu }{\nu } \right) ^{2} \frac{Q_{\nu }(A)}{1 - Q_{\nu }(A)}} &{} \text { if }\quad Q_{\nu }(A) \in [0,1) \\ 0 &{} \text { if }\quad Q_{\nu }(A)=1. \end{array}\right. } \end{aligned}$$

(17)

Moreover, as the $\nu$ parameter is the neutral value of the Dombi negation operator [see Eq. (12)], the following property of the $\nu$-additive measure holds as well.

Proposition 7

Let X be a finite set, $Q_{\nu }$ a $\nu$-additive measure on X and let the set $A_{\nu }$ be given as

$$\begin{aligned} A_{\nu } = \lbrace A \in \mathscr {P}(X) \vert Q_{\nu }(A) = \nu \rbrace , \end{aligned}$$

where $\nu \in (0, 1)$. Then, for any $A \in A_{\nu }$ the $Q_{\nu }$ measure of the complement set $\overline{A}$ is equal to $\nu$; that is, $Q_{\nu }(\overline{A}) = \nu$.

Proof

If $A \in A_{\nu }$, then $Q_{\nu }(A) = \nu$ and utilizing the $\nu$-additive negation given by Eq. (17), we have

$$\begin{aligned} Q_{\nu }(\overline{A}) = \frac{1}{1 + \left( \frac{1 - \nu }{\nu } \right) ^{2} \frac{\nu }{1 - \nu }} = \nu . \end{aligned}$$

$\square$

This result means that the $\nu$-additive complement operation may be viewed as a complement operation characterized by its fix point $\nu$.

3.2 Main properties of the $\nu$-additive ($\lambda$-additive) measures

It is worth mentioning that the definition of the $\nu$-additive measure is the same as that of the $\lambda$-additive measure with an alternative parameterization. Thus, utilizing the fact that any $\nu$-additive measure is a $\lambda$-additive measure with $\lambda = \left( \frac{1 - \nu }{\nu } \right) ^{2} -1$, some of the properties of $\lambda$-additive measures can be expressed in terms of $\nu$-additive measures and vice versa. In this section, we will discuss the main properties of these two measures. In many cases, to make the calculations simpler, we will use the $\lambda$-additive form to demonstrate some properties and then we will state them in terms of the $\nu$-additive measure as well. We will follow this approach from now on, and $Q_{\lambda }$ will always denote a $\lambda$-additive measure with the parameter $\lambda \in (-\,1, \infty )$ and $Q_{\nu }$ will always denote a $\nu$-additive measure with the parameter $\nu \in (0, 1)$.

3.2.1 $\nu$-additive ($\lambda$-additive) measure of collection of disjoint sets

Here, we will outline the computation of the $\nu$-additive ($\lambda$-additive) measure of collection of pairwise disjoint sets.

Proposition 8

If X is a finite set, $Q_{\lambda }$ is a $\lambda$-additive measure on X and $A_1, A_2, \ldots , A_n \in \mathscr {P}(X)$ are pairwise disjoint sets, then

$$\begin{aligned} \begin{aligned}&Q_{\lambda } \left( \bigcup \limits _{i=1}^n A_i \right) \\&\quad = {\left\{ \begin{array}{ll} \sum \limits _{i=1}^{n} Q_{\lambda }(A_i), &{} \text {if }\quad \lambda =0\\ \frac{1}{\lambda } \left( \prod \limits _{i=1}^{n} (1+\lambda Q_{\lambda }( A_i))-1 \right) , &{} \text {if }\quad \lambda >-1, \lambda \ne 0. \end{array}\right. } \end{aligned} \end{aligned}$$

(18)

Proof

Here, we will discuss the two possible cases: (1) $\lambda =0$, (2) $\lambda >-1$ and $\lambda \ne 0$.

• (1) In this case, the proposition trivially follows from the definition of the $\lambda$-additive measures.

• (2) Here, we will apply induction. By utilizing the definition of the $\lambda$-additive measures, the associativity of the union operation over sets and simple calculations, it can be shown that

  $$\begin{aligned} Q_{\lambda } \left( \bigcup \limits _{i=1}^n A_i \right) = \frac{1}{\lambda } \left( \prod \limits _{i=1}^{n} (1+\lambda Q_{\lambda }( A_i))-1 \right) \end{aligned}$$

  (19)

  holds for $n=2$ and $n=3$, where $A_{1}, A_{2}, A_{3} \in \mathscr {P}(X)$, $\lambda >-1$, $\lambda \ne 0$. Now, let us assume that Eq. (19) holds for any $A_{1}, A_{2}, \ldots , A_{n} \in \mathscr {P}(X)$, $\lambda >-1$, $\lambda \ne 0$. Let $G_{n}$ be defined as follows:

  $$\begin{aligned} G_{n} = \prod \limits _{i=1}^{n} (1+\lambda Q_{\lambda }( A_i)). \end{aligned}$$

  With this notation, $G_{n+1} = G_{n} \left( 1 + \lambda Q_{\lambda }(A_{n+1}) \right)$, and the equality that we seek to prove is

  $$\begin{aligned} Q_{\lambda } \left( \bigcup \limits _{i=1}^{n+1} A_i \right) = \frac{1}{\lambda } \left( G_{n+1}-1 \right) . \end{aligned}$$

  By utilizing the definition of the $\lambda$-additive measures and the associativity of the union operation over sets, we get

  $$\begin{aligned} \begin{aligned} Q_{\lambda } \left( \bigcup \limits _{i=1}^{n+1} A_i \right)&= Q_{\lambda } \left( \bigcup \limits _{i=1}^{n} A_i \right) + Q_{\lambda }(A_{n+1}) \\&\quad + \lambda Q_{\lambda } \left( \bigcup \limits _{i=1}^{n} A_i \right) Q_{\lambda }(A_{n+1}). \end{aligned} \end{aligned}$$

  Now, utilizing the inductive condition, the last equation can be written as

  $$\begin{aligned} \begin{aligned}&Q_{\lambda } \left( \bigcup \limits _{i=1}^{n+1} A_i \right) = \frac{1}{\lambda } \left( G_{n}-1 \right) + Q_{\lambda }(A_{n+1}) \\&\qquad + \lambda \frac{1}{\lambda } \left( G_{n}-1 \right) Q_{\lambda }(A_{n+1}) \\&\quad = \frac{1}{\lambda } \left( G_{n}-1 \right) \left( 1 + \lambda Q_{\lambda }(A_{n+1}) \right) + Q_{\lambda }(A_{n+1}) \\&\quad = \frac{1}{\lambda } G_{n} \left( 1 + \lambda Q_{\lambda }(A_{n+1}) \right) - \frac{1}{\lambda } \\&\quad = \frac{1}{\lambda } \left( G_{n} \left( 1 + \lambda Q_{\lambda }(A_{n+1}) \right) - 1 \right) = \frac{1}{\lambda } \left( G_{n+1}-1 \right) . \end{aligned} \end{aligned}$$

$\square$

Remark 5

Note that in Eq. (18), the case $\lambda =0$ may be viewed as a special case of $\lambda >-1$ and $\lambda \ne 0$. Namely, the right-hand side of Eq. (19) can be written as

$$\begin{aligned} \begin{aligned}&\frac{1}{\lambda } \left( \prod \limits _{i=1}^{n} (1+\lambda Q_{\lambda }( A_i))-1 \right) = \sum \limits _{i=1}^{n} Q_{\lambda }(A_i) \\&\qquad + \sum \limits _{k=2}^{n} \lambda ^{k-1} \sum \limits _{\begin{array}{c} 1 \le i_{1}< i_{2} \cdots \\ \cdots < i_{k} \le n \end{array}} Q_{\lambda }(A_{i_1}) Q_{\lambda }(A_{i_2}) \ldots Q_{\lambda }(A_{i_k}) \end{aligned} \end{aligned}$$

from which

$$\begin{aligned} \lim \limits _{\lambda \rightarrow 0} \left( \frac{1}{\lambda } \left( \prod \limits _{i=1}^{n} (1+\lambda Q_{\lambda }( A_i))-1 \right) \right) = \sum \limits _{i=1}^{n} Q_{\lambda }(A_i). \end{aligned}$$

Proposition 8 can be stated in terms of the $\nu$-additive measure as follows.

Proposition 9

If X is a finite set, $Q_{\nu }$ is a $\nu$-additive measure on X and $A_1, A_2, \ldots , A_n \in \mathscr {P}(X)$ are pairwise disjoint sets, then

$$\begin{aligned} \begin{aligned}&Q_{\nu } \left( \bigcup \limits _{i=1}^n A_i \right) \\&\quad = {\left\{ \begin{array}{ll} \sum \limits _{i=1}^{n} Q_{\nu }(A_i), &{} \text {if }\quad \nu = 1/2\\ y &{} \text {if }\quad \nu \ne 1/2, \end{array}\right. } \end{aligned} \end{aligned}$$

(20)

where $\nu \in (0,1)$,

$$\begin{aligned} y = \frac{1}{\left( \frac{1 - \nu }{\nu } \right) ^{2} -1 } \left( \prod \limits _{i=1}^{n} (1+ \left( \left( \frac{1 - \nu }{\nu } \right) ^{2} - 1 \right) Q_{\nu }( A_i))-1 \right) . \end{aligned}$$

Proof

Recalling Proposition 6, this proposition directly follows from Proposition 8. $\square$

3.2.2 General forms for the $\nu$-additive ($\lambda$-additive) measure of union and intersection of two sets

The calculations of the $\lambda$-additive measure and $\nu$-additive measure of two disjoint sets are given in Definitions 9 and 10, respectively. Here, we will show how the $\nu$-additive ($\lambda$-additive) measure of two sets can be computed when these sets are not disjoint. We will also discuss how the $\nu$-additive ($\lambda$-additive) measure of intersection of two sets can be computed.

Proposition 10

If X is a finite set and $Q_{\lambda }$ is a $\lambda$-additive measure on X, then for any $A, B \in \mathscr {P}(X)$,

$$\begin{aligned} \begin{aligned}&Q_{\lambda } ({A} \cup B) \\&\quad = \frac{Q_{\lambda } (A)+Q_{\lambda } (B) + \lambda Q_{\lambda } (A)Q_{\lambda } (B) - Q_{\lambda } (A \cap B) }{1+ \lambda Q_{\lambda } (A \cap B)}. \end{aligned} \end{aligned}$$

Proof

Since $A \cap (\overline{A} \cap B)=\emptyset$ and $A\cup (\overline{A} \cap B)= {A} \cup B$, applying Eq. (7) gives us

$$\begin{aligned} \begin{aligned}&Q_{\lambda } (A \cup B) \\&\quad = Q_{\lambda } (A)+Q_{\lambda } (\overline{A} \cap B) + \lambda Q_{\lambda } (A)Q_{\lambda } (\overline{A} \cap B) \\&\quad =Q_{\lambda } (A)+ Q_{\lambda } (\overline{A} \cap B)(1+\lambda Q_{\lambda } (A) ). \end{aligned} \end{aligned}$$

(21)

Next, since $(A \cap B) \cap (\overline{A}\cap B)=\emptyset$ and $(A\cap B)\cup (\overline{A} \cap B)= B$, applying Eq. (7) again gives

$$\begin{aligned} \begin{aligned} Q_{\lambda } (B)&= Q_{\lambda } ({A} \cap B)+Q_{\lambda } (\overline{A} \cap B) \\&\quad + \lambda Q_{\lambda } ({A} \cap B) Q_{\lambda } (\overline{A} \cap B) \\&=Q_{\lambda } ({A} \cap B)+ Q_{\lambda } (\overline{A} \cap B)(1+\lambda Q_{\lambda } (A \cap B) ). \end{aligned} \end{aligned}$$

(22)

Now, by expressing $Q_{\lambda }(\overline{A}\cap B)$ in terms of (22), we get

$$\begin{aligned} Q_{\lambda } (\overline{A} \cap B)=\frac{Q_{\lambda } (B) - Q_{\lambda } (A \cap B) }{ 1+ \lambda Q_{\lambda } (A \cap B)} \end{aligned}$$

and substituting this into (21), we get

$$\begin{aligned} \begin{aligned}&Q_{\lambda } ({A} \cup B) \\&\quad = Q_{\lambda } ({A})+\frac{Q_{\lambda } (B) - Q_{\lambda } (A \cap B) }{ 1+ \lambda Q_{\lambda } (A \cap B)}(1 +\lambda Q_{\lambda } ({A})) \\&\quad = \frac{Q_{\lambda } (A)+Q_{\lambda } (B) + \lambda Q_{\lambda } (A)Q_{\lambda } (B) - Q_{\lambda } (A \cap B) }{1+ \lambda Q_{\lambda } (A \cap B)}. \end{aligned} \end{aligned}$$

(23)

Hence, we have the general form of the $\lambda$-additive measure of the union of two sets. $\square$

Remark 6

Notice that if $\lambda =0$, then Eq. (23) reduces to $Q_{\lambda } ({A} \cup B) = Q_{\lambda } (A)+Q_{\lambda } (B) - Q_{\lambda } (A \cap B)$, which has the same form as the probability measure of union of two sets. Later, we will discuss how the $\lambda$-additive ($\nu$-additive) measure is related to the probability measure.

Remark 7

Note that Eq. (23) can be written in the following equivalent forms:

$$\begin{aligned} \begin{aligned}&Q_{\lambda } ({A} \cup B)+Q_{\lambda } ({A} \cap B)+\lambda Q_{\lambda } ({A} \cup B) Q_{\lambda } ({A} \cap B) \\&\quad = Q_{\lambda } ({A}) +Q_{\lambda } ( B)+ \lambda Q_{\lambda } ({A})Q_{\lambda } (B) \end{aligned} \end{aligned}$$

or for $\lambda \ne 0$

$$\begin{aligned} \begin{aligned}&\frac{1}{\lambda }\left( \left( 1+ Q_{\lambda } ({A} \cup B)\right) \left( 1 +\lambda Q_{\lambda } ({A} \cap B)\right) -1 \right) \\&\quad = \frac{1}{\lambda }\left( \left( 1+ Q_{\lambda } ({A})\right) \left( 1 +\lambda Q_{\lambda } (B)\right) -1 \right) . \end{aligned} \end{aligned}$$

Corollary 3

If X is a finite set and $Q_{\lambda }$ is a $\lambda$-additive measure on X, then for any $A, B \in \mathscr {P}(X)$,

$$\begin{aligned} \begin{aligned}&Q_{\lambda } ({A} \cap B) \\&\quad =\frac{Q_{\lambda } (A)+Q_{\lambda } (B) + \lambda Q_{\lambda } (A)Q_{\lambda } (B) - Q_{\lambda } (A \cup B) }{1+ \lambda Q_{\lambda } (A \cup B)}. \end{aligned} \end{aligned}$$

(24)

Proof

By expressing $Q_{\lambda } ({A} \cap B)$ in Eq. (23), we get Eq. (24). $\square$

3.2.3 Other properties of the $\nu$-additive ($\lambda$-additive) measure of union and the intersection of two sets

The following results are related to the $\nu$-additive measure of union and the intersection of two sets.

Proposition 11

Let X be a finite set, $Q_{\nu }$ be a $\nu$-additive measure on X and let $A, B \in \mathscr {P}(X)$. Then

• (1) if $A \cup B=X$ (complementing case), then

  $$\begin{aligned} \begin{aligned}&Q_{\nu } ({A} \cap B)= Q_{\nu }(A) Q_{\nu }(B) \\&\quad -\left( \frac{\nu }{1-\nu } \right) ^{2} (1- Q_{\nu }(A))(1-Q_{\nu }(B)) \end{aligned} \end{aligned}$$

  (25)

• (2) if $A \cap B=\emptyset$ (disjoint case), then

  $$\begin{aligned} \begin{aligned}&Q_{\nu } ({A} \cup B) \\&\quad =1- \Bigg ( (1-Q_{\nu }(A))(1-Q_{\nu }(B)) \\&\qquad - \left( \frac{1 - \nu }{\nu } \right) ^{2} Q_{\nu }(A) Q_{\nu }(B) \Bigg ). \end{aligned} \end{aligned}$$

  (26)

Proof

• (1) Since the $\nu$-additive measure $Q_{\nu }$ is identical to the $\lambda$-additive measure $Q_{\lambda }$ with $\lambda = \left( \frac{1 - \nu }{\nu } \right) ^{2} -1$, $Q_{\nu }({A} \cap B) = Q_{\lambda } ({A} \cap B)$. Now, utilizing the fact that $Q_{\nu }(A \cup B)$ = $Q_{\lambda }(A \cup B)$ = 1 and Eq. (24), $Q_{\lambda } ({A} \cap B)$ can be written as

  $$\begin{aligned} \begin{aligned} Q_{\lambda } ({A} \cap B)&= \frac{Q_{\lambda } (A)+Q_{\lambda } (B) + \lambda Q_{\lambda } (A)Q_{\lambda } (B) - 1}{1+ \lambda } \\&= \frac{(1+\lambda ) Q_{\lambda }(A) Q_{\lambda } (B)}{1+\lambda } \\&\quad - \frac{1-Q_{\lambda }(A) - Q_{\lambda }(B) + Q_{\lambda }(A)Q_{\lambda }(B)}{1+\lambda } \\&= Q_{\lambda }(A) Q_{\lambda }(B) - \frac{1}{1+\lambda } (1-Q_{\lambda }(A))(1-Q_{\lambda }(B)). \end{aligned} \end{aligned}$$

  And by using the equation $\lambda = \left( \frac{1 - \nu }{\nu } \right) ^{2} -1$, we get

  $$\begin{aligned} \begin{aligned}&Q_{\nu } ({A} \cap B) \\&\quad = Q_{\nu }(A) Q_{\nu }(B) - \left( \frac{\nu }{1-\nu } \right) ^{2} (1- Q_{\nu }(A))(1-Q_{\nu }(B)). \end{aligned} \end{aligned}$$

• (2) Since $A \cap B = \emptyset$, applying the definition of the $\nu$-additive measure gives

  $$\begin{aligned} \begin{aligned}&Q_{\nu }(A \cup B) \\&\quad = Q_{\nu }(A)+Q_{\nu }(B) + \left( \left( \frac{1 - \nu }{\nu } \right) ^{2}-1 \right) Q_{\nu }(A)Q_{\nu }(B) \\&\quad = 1-(1-Q_{\nu }(A) - Q_{\nu }(B) + Q_{\nu }(A)Q_{\nu }(B)) \\&\qquad + \left( \frac{1 - \nu }{\nu } \right) ^{2} Q_{\nu }(A)Q_{\nu }(B) \\&\quad = 1-\Bigg ( (1-Q_{\nu }(A))(1-Q_{\nu }(B)) \\&\qquad - \left( \frac{1 - \nu }{\nu } \right) ^{2} Q_{\nu }(A) Q_{\nu }(B) \Bigg ). \end{aligned} \end{aligned}$$

$\square$

Note that the term $\left( \frac{\nu }{1-\nu } \right) ^{2} (1- Q_{\nu }(A))(1-Q_{\nu }(B))$ in Eq. (25) may be regarded as the corrective term of the intersection; that is, if $\nu \rightarrow 0$, then $Q_{\nu } ({A} \cap B)= Q_{\nu }(A) Q_{\nu }(B)$. Similarly, the term $\left( \frac{1 - \nu }{\nu } \right) ^{2} Q_{\nu }(A) Q_{\nu }(B)$ in Eq. (26) may be interpreted as the corrective term of the union; that is, if $\nu \rightarrow 1$, then $Q_{\nu }(A \cup B) = 1-(1-Q_{\nu }(A))(1-Q_{\nu }(B))$.

3.2.4 Characterization by independent variables

We have demonstrated (see Proposition 8) that if X is a finite set, $Q_{\lambda }$ is a $\lambda$-additive measure on X, $\lambda >-1$, $\lambda \ne 0$, $A_1, A_2, \ldots , A_n \in \mathscr {P}(X)$ are pairwise disjoint sets, and

$$\begin{aligned} A = \bigcup \limits _{i=1}^n A_i, \end{aligned}$$

then

$$\begin{aligned} Q_{\lambda }(A) = Q_{\lambda } \left( \bigcup \limits _{i=1}^n A_i \right) = \frac{1}{\lambda } \left( \prod \limits _{i=1}^{n} (1+\lambda Q_{\lambda }(A_i))-1 \right) . \end{aligned}$$

It means that the value of $Q_{\lambda }(A)$ can be readily calculated from the independent values $Q_{\lambda }({A_i})$, where $i = 1, 2, \ldots , n$. If $X = \lbrace A_1, A_2, \ldots , A_n \rbrace$, then

$$\begin{aligned} \begin{aligned} Q_{\lambda }(X)&= Q_{\lambda } \left( \bigcup \limits _{i=1}^n A_i \right) \\&= \frac{1}{\lambda } \left( \prod \limits _{i=1}^{n} (1+\lambda Q_{\lambda }(A_i))-1 \right) =1. \end{aligned} \end{aligned}$$

(27)

The following proposition demonstrates that Eq. (27) has only one root in the interval $(-1,0) \cup (0,\infty )$.

Proposition 12

If X is a finite set, $Q_{\lambda }$ is a $\lambda$-additive measure on X, $\lambda >-1$, $\lambda \ne 0$, $A_1, A_2, \ldots , A_n \in \mathscr {P}(X)$ are pairwise disjoint sets such that $Q_{\lambda }(A_i)<1$, $i \in \lbrace 1,2, \ldots , n \rbrace$, then the equation

$$\begin{aligned} \frac{1}{\lambda } \left( \prod \limits _{i=1}^{n} (1+\lambda Q_{\lambda }(A_i))-1 \right) =1 \end{aligned}$$

(28)

has only one root in the interval $(-\,1,0) \cup (0,\infty )$.

Proof

This proof is based on the proof of a theorem connected with the multiplicative utility functions described by Keeney (1974, Appendix B). Since $\lambda \ne 0$, Eq. (28) can be written as

$$\begin{aligned} \lambda + 1 = \prod \limits _{i=1}^{n} (1+\lambda z_i), \end{aligned}$$

(29)

where $z_i=Q_{\lambda }(A_i)$, $i=1,2, \ldots , n$. Now, let $S=\sum _{i=1}^{n} z_i$ and let the polynomial f(q) be given as

$$\begin{aligned} f(q) = q + 1 - \prod \limits _{i=1}^{n} (1+q z_i), \end{aligned}$$

(30)

where $-1 \le q < \infty$. From Eqs. (29) and (30), we get the following results:

$$\begin{aligned} f(\lambda )=0, \text { } f(0)=0, \text { } f(-1) = - \prod \limits _{i=1}^{n} (1 - z_i) < 0. \end{aligned}$$

The first derivative of function f is

$$\begin{aligned} f'(q) = \frac{\mathrm {d}f(q)}{\mathrm {d}q} = 1- \sum \limits _{i=1}^{n} z_i \prod \limits _{i \ne j} (1+z_j q), \end{aligned}$$

from which we can see that $f'(q)$ is decreasing (with respect to q) in the interval $(-\,1,\infty )$,

$$\begin{aligned} f'(0) = 1 - \sum _{i=1}^{n} z_i = 1-S \end{aligned}$$

(31)

and

$$\begin{aligned} \lim \limits _{q \rightarrow \infty } f'(q) = - \infty . \end{aligned}$$

(32)

Here, we will distinguish three cases: (1) $S<1$; (2) $S=1$; (3) $S>1$.

• (1) Equation (31) implies that if $S<1$, then $f'(0)>0$. Since $f'(0)>0$ and $f'(q)$ is decreasing in the interval $(-\,1,\infty )$, $f'(q)$ is positive in $(-\,1, 0)$. Therefore, $f'(q) = 0$ has no root in $(-\,1,0)$. Based on Eq. (32), $f'(\infty ) = -\infty$, and so $f'(q) = 0$ has a unique root $q^{*}$ in $(0,\infty )$. Since $f(0) = 0$ and $f'(q) > 0$ in $(0, q^{*})$, $f(q) = 0$ has no root in $(0, q^{*})$. As $f(q^{*})>0$ and $f'(q)$ is negative and decreasing to $-\infty$ in $(q^{*},\infty )$, $f(q)=0$ has a unique root $q_0$ in $(q^{*},\infty )$. Moreover, $f(q)>0$ in $(0,q_0)$ and $f(q)<0$ in $(q_0, \infty )$; that is, the unique root $q_0$ is in $(0, \infty )$.

• (2) It follows from Eq. (31) that if $S=1$, then $f'(0)=0$. Since $f'(0)=0$ and $f'(q)$ is decreasing in the interval $(-\,1,\infty )$, $f'(q)$ is positive in the interval $(-\,1,0)$ and it is negative in the interval $(0, \infty )$. Thus, $q=0$ is the only root of $f'(q)=0$ in the interval $(-\,1,\infty )$. Moreover, since $f(0)=0$, $q=0$ is the only root of $f(q)=0$. It means that if $S=1$, then the only solution of Eq. (29) is $\lambda =0$. Recall that $\lambda \ne 0$; that is, in this case we do not get any solution to the equation in (28).

• (3) Equation (31) implies that if $S>1$, then $f'(0)<0$. Since $f'(0)<0$ and $f'(q)$ is decreasing in the interval $(-\,1,\infty )$, $f'(q)$ is negative in $(0,\infty )$. As $f(0)=0$ and $f'(q)$ is negative in $(0,\infty )$, $f(q)=0$ has no root in $(0,\infty )$. On the one hand, as $f(0)=0$ and $f'(0)<0$, $f(q)>0$ immediately to the left of zero. On the other hand, $f(-\,1) < 0$. It means that there must be at least one root $q_0$ of $f(q)=0$ in $(-\,1,0)$. Since $f'(q)$ is decreasing and $f(0)=0$, $q_0$ is the unique root of $f(q)=0$ in (0, 1). $\square$

Proposition 12 tells us that Eq. (27) can be solved numerically for $\lambda$ in the interval $(-\,1, 0)$ or in the interval $(0, \infty )$. Hence, the $\lambda$-additive measure $Q_{\lambda }$ can be unambiguously characterized by n independent variables.

3.3 Dual $\nu$-additive ($\lambda$-additive) measures and their properties

Later, we will utilize the concept of the dual pair of $\lambda$-additive measures and the concept of the dual pair of $\nu$-additive measures.

Definition 11

Let $Q_{\lambda _1}$ and $Q_{\lambda _2}$ be two $\lambda$-additive measures on the finite set X. Then, $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are said to be a dual pair of $\lambda$-additive measures iff

$$\begin{aligned} Q_{\lambda _1} ({A}) +Q_{\lambda _2} (\overline{A})=1 \end{aligned}$$

holds for any $A \in \mathscr {P}(X)$.

Definition 12

Let $Q_{\nu _1}$ and $Q_{\nu _2}$ be two $\nu$-additive measures on the finite set X. Then, $Q_{\nu _1}$ and $Q_{\nu _2}$ are said to be a dual pair of $\nu$-additive measures iff

$$\begin{aligned} Q_{\nu _1} ({A}) +Q_{\nu _2} (\overline{A})=1 \end{aligned}$$

holds for any $A \in \mathscr {P}(X)$.

Later, we will utilize the following proposition.

Proposition 13

Let $Q_{\lambda _1}$ and $Q_{\lambda _2}$ be two $\lambda$-additive measures on the finite set X and let

$$\begin{aligned} \lambda _2 = - \frac{\lambda _1}{1+\lambda _1}. \end{aligned}$$

(33)

Then, for any $A \in \mathscr {P}(X)$

$$\begin{aligned} Q_{\lambda _2} ({A}) > 1- Q_{\lambda _1} (\overline{A}), \end{aligned}$$

(34)

if and only if

$$\begin{aligned} Q_{\lambda _2} ({\overline{A}}) < 1- Q_{\lambda _1} (A). \end{aligned}$$

(35)

Proof

Firstly, we will show that if $\lambda _2 = - \frac{\lambda _1}{1+\lambda _1}$ and $Q_{\lambda _2} ({A}) > 1- Q_{\lambda _1} (\overline{A})$ holds for any $A \in \mathscr {P}(X)$, then $Q_{\lambda _2} ({\overline{A}}) < 1- Q_{\lambda _1} (A)$ holds as well. By utilizing the formula for the $\lambda$-additive measure of complementer set given by Eq. (8), we get

$$\begin{aligned} Q_{\lambda _2}(A) = \frac{1-Q_{\lambda _2}({\overline{A}})}{1+\lambda _2 Q_{\lambda _2}({\overline{A}})} \end{aligned}$$

and

$$\begin{aligned} Q_{\lambda _1}(\overline{A}) = \frac{1-Q_{\lambda _1}(A)}{1+\lambda _1 Q_{\lambda _1}(A)} \end{aligned}$$

for any $A \in \mathscr {P}(X)$. Next, based on the condition $Q_{\lambda _2} ({A}) > 1- Q_{\lambda _1} (\overline{A})$, we have the following inequality:

$$\begin{aligned} \frac{1-Q_{\lambda _2}({\overline{A}})}{1+\lambda _2 Q_{\lambda _2}({\overline{A}})} > 1- \frac{1-Q_{\lambda _1}(A)}{1+\lambda _1 Q_{\lambda _1}(A)}. \end{aligned}$$

(36)

From Eq. (36), via simple calculations, we get

$$\begin{aligned} \begin{aligned}&1-Q_{\lambda _1}(A) - Q_{\lambda _2}({\overline{A}}) \\&\quad > Q_{\lambda _1}(A) Q_{\lambda _2}({\overline{A}}) (\lambda _1+\lambda _2+ \lambda _1 \lambda _2). \end{aligned} \end{aligned}$$

(37)

From the condition $\lambda _2 = - \frac{\lambda _1}{1+\lambda _1}$, we have the equation $\lambda _1+\lambda _2+ \lambda _1 \lambda _2 = 0$, and so the inequality relation in Eq. (37) can be written as

$$\begin{aligned} 1-Q_{\lambda _1}(A) - Q_{\lambda _2}({\overline{A}}) > 0, \end{aligned}$$

which is equivalent to that stated in Eq. (35).

Secondly, we will show that if $\lambda _2 = - \frac{\lambda _1}{1+\lambda _1}$ and $Q_{\lambda _2} ({\overline{A}}) < 1- Q_{\lambda _1} (A)$ holds for any $A \in \mathscr {P}(X)$, then $Q_{\lambda _2} ({A}) > 1- Q_{\lambda _1} (\overline{A})$ holds as well. By utilizing the formula for the $\lambda$-additive measure of complementer set given by Eq. (8), we get

$$\begin{aligned} Q_{\lambda _2}(\overline{A}) = \frac{1-Q_{\lambda _2}(A)}{1+\lambda _2 Q_{\lambda _2}(A)} \end{aligned}$$

and

$$\begin{aligned} Q_{\lambda _1}(A) = \frac{1-Q_{\lambda _1}(\overline{A})}{1+\lambda _1 Q_{\lambda _1}(\overline{A})} \end{aligned}$$

for any $A \in \mathscr {P}(X)$. Next, based on the condition $Q_{\lambda _2} ({\overline{A}}) < 1- Q_{\lambda _1} (A)$, we have the following inequality:

$$\begin{aligned} \frac{1-Q_{\lambda _2}(A)}{1+\lambda _2 Q_{\lambda _2}(A)} < 1- \frac{1-Q_{\lambda _1}(\overline{A})}{1+\lambda _1 Q_{\lambda _1}(\overline{A})}. \end{aligned}$$

(38)

From Eq. (38), by direct calculations, we get

$$\begin{aligned} \begin{aligned}&1-Q_{\lambda _1}(\overline{A}) - Q_{\lambda _2}(A) \\&\quad < Q_{\lambda _1}(\overline{A}) Q_{\lambda _2}(A) (\lambda _1+\lambda _2+ \lambda _1 \lambda _2). \end{aligned} \end{aligned}$$

(39)

Since the condition $\lambda _2 = - \frac{\lambda _1}{1+\lambda _1}$ is equivalent to the equation $\lambda _1+\lambda _2+ \lambda _1 \lambda _2 = 0$, the inequality relation in Eq. (39) can be written as

$$\begin{aligned} 1-Q_{\lambda _1}(\overline{A}) - Q_{\lambda _2}(A) <0, \end{aligned}$$

which is equivalent to that stated in Eq. (34). $\square$

Here, we will demonstrate some key properties of the $\nu$-additive ($\lambda$-additive) measure related to a dual pair of $\nu$-additive ($\lambda$-additive) measures.

Proposition 14

Let $Q_{\lambda _1}$ and $Q_{\lambda _2}$ be two $\lambda$-additive measures on the finite set X. Then, $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are a dual pair of $\lambda$-additive measures if and only if

$$\begin{aligned} \lambda _2 = - \frac{\lambda _1}{1+\lambda _1}. \end{aligned}$$

Proof

Firstly, we will show that if $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are a dual pair of $\lambda$-additive measures on the finite set X, then $\lambda _2 = - \frac{\lambda _1}{1+\lambda _1}$. Let $Q_{\lambda _1}$ and $Q_{\lambda _2}$ be a dual pair of $\lambda$-additive measures on X. It means that $Q_{\lambda _2} ({A}) = 1 - Q_{\lambda _1}(\overline{A})$ holds for any $A \in \mathscr {P}(X)$. Next, let $A,B \in \mathscr {P}(X)$ such that $A \cap B=\emptyset$. Then, $X=\overline{A \cap B} = \overline{A} \cup \overline{B}$. Now, noting that $Q_{\lambda _2} ({A}) = 1 - Q_{\lambda _1}(\overline{A})$, the formula for the $\lambda$-additive measure of the intersection of two sets given by Eq. (24) and the fact that $Q_{\lambda _1}(\overline{A} \cup \overline{B}) = Q_{\lambda _1}(X)=1$, we get

$$\begin{aligned} \begin{aligned} Q_{\lambda _2}(A \cup B)&= 1- Q_{\lambda _1}(\overline{A \cup B}) \\&= 1- Q_{\lambda _1}(\overline{A} \cap \overline{B}) = 1- \frac{Q_{\lambda _1}(\overline{A}) + Q_{\lambda _1}(\overline{B})}{1+ \lambda _1 Q_{\lambda _1}(\overline{A} \cup \overline{B})} \\&\quad -\frac{\lambda _1 Q_{\lambda _1}(\overline{A}) Q_{\lambda _1}(\overline{B}) - Q_{\lambda _1}(\overline{A} \cup \overline{B})}{1+ \lambda _1 Q_{\lambda _1}(\overline{A} \cup \overline{B})} \\&= 1- \frac{1-Q_{\lambda _2}(A) + 1-Q_{\lambda _2}(B)}{1+ \lambda _1} \\&- \frac{\lambda _1 (1-Q_{\lambda _2}(A))(1-Q_{\lambda _2}(B)) - 1}{1+ \lambda _1} \\&= Q_{\lambda _2}(A) + Q_{\lambda _2}(B) - \frac{\lambda _1}{1+\lambda _1} Q_{\lambda _2}(A) Q_{\lambda _2}(B). \end{aligned} \end{aligned}$$

(40)

Moreover, since $Q_{\lambda _2}$ is a $\lambda$-additive measure and $A \cap B = \emptyset$, the equation

$$\begin{aligned} \begin{aligned} Q_{\lambda _2}(A \cup B) = Q_{\lambda _2}(A) + Q_{\lambda _2}(B) + \lambda _2 Q_{\lambda _2}(A) Q_{\lambda _2}(B) \end{aligned} \end{aligned}$$

(41)

holds. Thus, from Eqs. (40) and (41) we get that $\lambda _2 = - \frac{\lambda _1}{1+\lambda _1}$.

Secondly, we will show that if $\lambda _2 = - \frac{\lambda _1}{1+\lambda _1}$, then $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are a dual pair of $\lambda$-additive measures on X. Let $\lambda _2 = - \frac{\lambda _1}{1+\lambda _1}$. Here, we seek to show that $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are a dual pair of $\lambda$-additive measures; that is, $Q_{\lambda _2} ({A}) = 1 - Q_{\lambda _1}(\overline{A})$ holds for any $A \in \mathscr {P}(X)$. Now, we will give an indirect proof of this. Let us assume that $\lambda _2 = - \frac{\lambda _1}{1+\lambda _1}$, but either (1) $Q_{\lambda _2} ({A}) > 1 - Q_{\lambda _1}(\overline{A})$, or (2) $Q_{\lambda _2} ({A}) < 1 - Q_{\lambda _1}(\overline{A})$ holds for any $A \in \mathscr {P}(X)$. We will show that this assumption leads to contradictions. Let $A,B \in \mathscr {P}(X)$ such that $A \cap B=\emptyset$. Then, $X=\overline{A \cap B} = \overline{A} \cup \overline{B}$.

• (1) Here, as $Q_{\lambda _2} ({A}) > 1 - Q_{\lambda _1}(\overline{A})$ holds for any $A \in \mathscr {P}(X)$, $Q_{\lambda _1}(\overline{A}) > 1- Q_{\lambda _2}({A})$ holds as well, and applying it to $\overline{A} \cap \overline{B}$, we get

  $$\begin{aligned} Q_{\lambda _1}(A \cup B) = Q_{\lambda _1}(\overline{\overline{A} \cap \overline{B}}) > 1- Q_{\lambda _2}(\overline{A} \cap \overline{B}). \end{aligned}$$

  (42)

  Utilizing the formula for the $\lambda$-additive measure of the intersection of two sets given by Eq. (24) and the fact that $Q_{\lambda _2}(\overline{A} \cup \overline{B}) = Q_{\lambda _2}(X)=1$, the right-hand side of Eq. (42) can be expressed as

  $$\begin{aligned} \begin{aligned}&1- Q_{\lambda _2}(\overline{A} \cap \overline{B}) \\&\quad = 1- \frac{Q_{\lambda _2}(\overline{A}) + Q_{\lambda _2}(\overline{B}) + \lambda _2 Q_{\lambda _2}(\overline{A}) Q_{\lambda _2}(\overline{B}) - 1}{1+ \lambda _2}. \end{aligned} \end{aligned}$$

  (43)

  Now, applying the inequality $Q_{\lambda _2} ({A}) > 1 - Q_{\lambda _1}(\overline{A})$ to set B, we have $Q_{\lambda _2} ({B}) > 1 - Q_{\lambda _1}(\overline{B})$, and so utilizing the fact that $\lambda _2 = - \frac{\lambda _1}{1+\lambda _1}$, $Q_{\lambda _2} ({A}) > 1 - Q_{\lambda _1}(\overline{A})$ and $Q_{\lambda _2} ({B}) > 1 - Q_{\lambda _1}(\overline{B})$, Proposition 13 yields the inequality relations

  $$\begin{aligned} Q_{\lambda _2} ({\overline{A}}) < 1- Q_{\lambda _1} (A) \end{aligned}$$

  (44)

  and

  $$\begin{aligned} Q_{\lambda _2} ({\overline{B}}) < 1- Q_{\lambda _1} (B). \end{aligned}$$

  (45)

  Next, noting Eqs. (44) and (45), from Eq. (43) we can further derive the result

  $$\begin{aligned} \begin{aligned}&1- \frac{Q_{\lambda _2}(\overline{A}) + Q_{\lambda _2}(\overline{B}) + \lambda _2 Q_{\lambda _2}(\overline{A}) Q_{\lambda _2}(\overline{B}) - 1}{1+ \lambda _2} \\&\quad > 1- \frac{1-Q_{\lambda _1}(A) + 1-Q_{\lambda _1}(B)}{1+ \lambda _2} \\&\qquad - \frac{\lambda _2 (1-Q_{\lambda _1}(A))(1-Q_{\lambda _1}(B)) - 1}{1+ \lambda _2} \\&\quad = Q_{\lambda _1}(A) + Q_{\lambda _1}(B) - \frac{\lambda _2}{1+\lambda _2} Q_{\lambda _1}(A) Q_{\lambda _1}(B). \end{aligned} \end{aligned}$$

  (46)

  On the one hand, utilizing $\lambda _2 = - \frac{\lambda _1}{1+\lambda _1}$, from Eqs. (42), (43) and (46), we get

  $$\begin{aligned} Q_{\lambda _1}(A \cup B) > Q_{\lambda _1}(A) + Q_{\lambda _1}(B) + \lambda _1 Q_{\lambda _1}(A) Q_{\lambda _1}(B). \end{aligned}$$

  On the other hand, as $A \cap B=\emptyset$ and $Q_{\lambda _1}$ is a $\lambda$-additive measure, we have

  $$\begin{aligned} Q_{\lambda _1}(A \cup B) = Q_{\lambda _1}(A) + Q_{\lambda _1}(B) + \lambda _1 Q_{\lambda _1}(A) Q_{\lambda _1}(B). \end{aligned}$$

  Thus, the assumption that $\lambda _2 = - \frac{\lambda _1}{1+\lambda _1}$ and $Q_{\lambda _2} ({A}) > 1 - Q_{\lambda _1}(\overline{A})$ leads to a contradiction.

• (2) Following the same steps as in case (1), the assumption that $\lambda _2 = - \frac{\lambda _1}{1+\lambda _1}$ and $Q_{\lambda _2} ({A}) < 1 - Q_{\lambda _1}(\overline{A})$ leads to the inequality

  $$\begin{aligned} Q_{\lambda _1}(A \cup B) < Q_{\lambda _1}(A) + Q_{\lambda _1}(B) + \lambda _1 Q_{\lambda _1}(A) Q_{\lambda _1}(B), \end{aligned}$$

  which contradicts the fact that $Q_{\lambda _1}$ is a $\lambda$-additive measure.

Based on case (1) and case (2), we may conclude that assuming that $\lambda _2 = - \frac{\lambda _1}{1+\lambda _1}$ and $Q_{\lambda _2} ({A}) \ne 1 - Q_{\lambda _1}(\overline{A})$ leads to contradictions. That is, we have proven that if $\lambda _2 = - \frac{\lambda _1}{1+\lambda _1}$, then $Q_{\lambda _2} ({A}) = 1 - Q_{\lambda _1}(\overline{A})$ holds. It means that the equation $\lambda _2 = - \frac{\lambda _1}{1+\lambda _1}$ implies that $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are a dual pair of $\lambda$-additive measures on X. $\square$

Proposition 14 can be stated in terms of the $\nu$-additive measure as follows.

Proposition 15

Let $Q_{\nu _1}$ and $Q_{\nu _2}$ be two $\nu$-additive measures on the finite set X. Then, $Q_{\nu _1}$ and $Q_{\nu _2}$ are a dual pair of $\nu$-additive measures if and only if

$$\begin{aligned} \nu _1 + \nu _2 =1. \end{aligned}$$

Proof

Utilizing Proposition 6, this proposition immediately follows from Proposition 14. $\square$

Utilizing the definition of the dual pair of $\lambda$-additive measures, the following corollary can be stated.

Corollary 4

Let $Q_{\lambda _1}$ and $Q_{\lambda _2}$ be a dual pair of $\lambda$-additive measures on the finite set X. Then, $\lambda _1 \in (-\,1, 0]$ if and only if $\lambda _2 \in [ 0, \infty )$.

Proof

Since $\lambda _2=-\frac{\lambda _1}{1+\lambda _1}$ is a bijection from $(-\,1, 0]$ to $[0, \infty )$, this corollary follows from Proposition 14. $\square$

Corollary 4 can be stated in terms of the $\nu$-additive measure as follows.

Corollary 5

Let $Q_{\nu _1}$ and $Q_{\nu _2}$ be a dual pair of $\nu$-additive measures on the finite set X. Then, $\nu _1 \in [ 1/2, 1)$ if and only if $\nu _2 \in (0, 1/2]$.

Proof

Taking into account Proposition 6, this corollary immediately follows from Corollary 4. $\square$

It should be mentioned here that one of the $\lambda$ parameters of a dual pair of $\lambda$-additive measures is always in the unbounded interval $[0,\infty )$. At the same time, the $\nu$ parameters of a dual pair of $\nu$-additive measures are both in a bounded interval; namely, one of them is in the interval (0, 1 / 2] and the other one is in the interval [1 / 2, 1).

3.3.1 The decomposition property of the $\lambda$-additive measure

The following proposition reveals an interesting property of the $\lambda$-additive measures.

Proposition 16

If X is a finite set and $Q_{\lambda }$ is a $\lambda$-additive measure on X, $A_1, \ldots , A_n, B_1, \ldots , B_m \in \mathscr {P}(X)$, $A_i \cap A_j=\emptyset$, $B_i \cap B_j=\emptyset$ for all $i \ne j$, $A_i \cap B_j=\emptyset$ for all i, j, $\lambda >-\,1$, $\lambda \ne 0$ and

$$\begin{aligned} A = \bigcup \limits _{i=1}^n A_i; B = \bigcup \limits _{i=1}^m B_i, \end{aligned}$$

then

$$\begin{aligned} \begin{aligned}&Q_{\lambda }(A \cup B) \\&\quad = \frac{1}{\lambda } \left( \prod \limits _{i=1}^n\left( 1+ \lambda Q_{\lambda }(A_i) \right) \prod \limits _{i=1}^m\left( 1+ \lambda Q_{\lambda }(B_i) \right) -\,1 \right) . \end{aligned} \end{aligned}$$

Proof

Since A and B are two disjoint sets and $Q_{\lambda }$ is a $\lambda$-additive measure,

$$\begin{aligned} Q_{\lambda }(A \cup B) = Q_{\lambda }(A) + Q_{\lambda }(B) + \lambda Q_{\lambda }(A) Q_{\lambda }(B) \end{aligned}$$

holds $(\lambda >-\,1, \lambda \ne 0)$. Next, utilizing the conditions $\lambda \ne 1$, $A_i \cap A_j=\emptyset$, $B_i \cap B_j=\emptyset$ for all $i \ne j$,

$$\begin{aligned} A = \bigcup \limits _{i=1}^n A_i; B = \bigcup \limits _{i=1}^m B_i, \end{aligned}$$

and the result of Proposition 8, $Q_{\lambda }(A \cup B)$ can be written as

$$\begin{aligned} \begin{aligned}&Q_{\lambda }(A \cup B) \\&\quad = \frac{1}{\lambda }(G-\,1) + \frac{1}{\lambda }(H-\,1) +\lambda \frac{1}{\lambda }(G-\,1) \frac{1}{\lambda }(H-\,1), \end{aligned} \end{aligned}$$

(47)

where

$$\begin{aligned} \begin{aligned} G&= \prod \limits _{i=1}^n \left( 1+ \lambda Q_{\lambda }(A_i) \right) \\ H&= \prod \limits _{i=1}^m \left( 1+ \lambda Q_{\lambda }(B_i) \right) . \end{aligned} \end{aligned}$$

Here, Eq. (47) can be written as

$$\begin{aligned} \begin{aligned}&Q_{\lambda }(A \cup B) \\&\quad = \frac{1}{\lambda }(G-1) + \frac{1}{\lambda }(H-\,1) + \frac{1}{\lambda }(G-1) (H-1) \\&\quad = \frac{1}{\lambda } (GH-1). \end{aligned} \end{aligned}$$

Hence,

$$\begin{aligned} \begin{aligned}&Q_{\lambda }(A \cup B) \\&\quad = \frac{1}{\lambda } \left( \prod \limits _{i=1}^n\left( 1+ \lambda Q_{\lambda }(A_i) \right) \prod \limits _{i=1}^m\left( 1+ \lambda Q_{\lambda }(B_i) \right) -\,1 \right) . \end{aligned} \end{aligned}$$

$\square$

4 Connection with belief-, probability- and plausibility measures

Here, we will discuss some important properties of the $\nu$-additive ($\lambda$-additive) measure and how it is connected to the belief-, probability- and plausibility measures.

Proposition 17

Let X be a finite set and let $Q_{\lambda }$ be a $\lambda$-additive measure on X. Then, on set X, $Q_{\lambda }$ is a

• (1) plausibility measure if and only if $-\,1< \lambda \le 0$

• (2) probability measure if and only if $\lambda =0$

• (3) belief measure if and only if $\lambda \ge 0$.

Proof

See Dubois and Prade (1980) and Banon (1978) $\square$

Note that in terms of the $\nu$-additive measure, Proposition 17 can be stated as follows.

Proposition 18

Let X be a finite set and let $Q_{\nu }$ be a $\nu$-additive measure on X. Then, on set X, $Q_{\nu }$ is a

• (1) belief measure if and only if $0<\nu \le 1/2$

• (2) probability measure if and only if $\nu =1/2$

• (3) plausibility measure if and only if $1/2 \le \nu <1$.

Proof

Taking into account Proposition 6, this proposition immediately follows from Proposition 17. $\square$

Fig. 1

$\nu$-additive measures of set A versus $\nu$-additive measures of complement of A

Figure 1 shows the connection between $Q_{\nu }(\overline{A})$ and $Q_{\nu }(A)$ for various values of parameter $\nu$ of the $\nu$-additive measure $Q_{\nu }$. From this figure, in accordance with Proposition 18, we notice the following. If $\nu =1/2$, then $Q_{\nu }$ is a probability measure and so $Q_{\nu }(\overline{A}) = 1- Q_{\nu }(A)$. If $0<\nu \le 1/2$, then $Q_{\nu }$ is a belief measure and $Q_{\nu }(\overline{A}) \le 1- Q_{\nu }(A)$. If $1/2 \le \nu <1$, then $Q_{\nu }$ is a plausibility measure and $Q_{\nu }(\overline{A}) \ge 1- Q_{\nu }(A)$. Moreover, in accordance with Eq. (17), for a given set A, $Q_{\nu }(\overline{A})$ increases with the value of parameter $\nu$. That is, the smaller the value of parameter $\nu$, the stronger the complement operation. It also means that any belief measure of a complement set is always less than or equal to any plausibility measure of the same complement set.

Proposition 19

Let $Q_{\lambda _1}$ and $Q_{\lambda _2}$ be two $\lambda$-additive measures on the finite set X. Then, $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are a dual pair of belief- and plausibility measures on X if and only if they are a dual pair of $\lambda$-additive measures on X.

Proof

Firstly, we will show that if the condition of the proposition is satisfied and $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are a dual pair of belief- and plausibility measures on X, then $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are a dual pair of $\lambda$-additive measures on X. Let $Q_{\lambda _1}$ and $Q_{\lambda _2}$ be a dual pair of belief- and plausibility measures on X. Since, $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are a dual pair; that is, $Q_{\lambda _2}(A) = 1-Q_{\lambda _1}(\overline{A})$ holds for any $A \in \mathscr {P}(X)$, and $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are $\lambda$-additive measures on X, they are also a dual pair of $\lambda$-additive measures on X.

Secondly, we will show that if $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are a dual pair of $\lambda$-additive measures on X, then $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are a dual pair of belief- and plausibility measures on X. Let $Q_{\lambda _1}$ and $Q_{\lambda _2}$ be a dual pair of $\lambda$-additive measures on X. Then, based on Corollary 4, either $\lambda _1 \in (-\,1, 0]$ and $\lambda _2 \in [ 0, \infty )$, or $\lambda _1 \in [ 0, \infty )$ and $\lambda _2 \in (-\,1, 0]$ holds. Now, utilizing Proposition 17, we get that either $Q_{\lambda _1}$ is a plausibility measure and $Q_{\lambda _2}$ is a belief measure, or $Q_{\lambda _1}$ is a belief measure and $Q_{\lambda _2}$ is a plausibility measure. Thus, noting that $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are a dual pair of $\lambda$-additive measures on X, we may conclude that they are also a dual pair of belief- and plausibility measures on X. $\square$

Proposition 19 can be stated in terms of the $\nu$-additive measure as follows.

Proposition 20

Let $Q_{\nu _1}$ and $Q_{\nu _2}$ be two $\nu$-additive measures on the finite set X. Then, $Q_{\nu _1}$ and $Q_{\nu _2}$ are a dual pair of belief- and plausibility measures on X if and only if they are a dual pair of $\nu$-additive measures on X.

Proof

Taking into account Proposition 6, this proposition directly follows from Proposition 19. $\square$

Proposition 21

Let $Q_{\lambda _1}$ and $Q_{\lambda _2}$ be two $\lambda$-additive measures on the finite set X. Then, $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are a dual pair of belief- and plausibility measures on X if and only if

$$\begin{aligned} \lambda _2 = -\frac{\lambda _1}{1+\lambda _1}. \end{aligned}$$

Proof

Following Proposition 19, if $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are two $\lambda$-additive measures on the finite set X, then $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are a dual pair of belief- and plausibility measures on X if and only if they are a dual pair of $\lambda$-additive measures on X. Furthermore, based on Proposition 14, if $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are two $\lambda$-additive measures on the finite set X, then $Q_{\lambda _1}$ and $Q_{\lambda _2}$ are a dual pair of $\lambda$-additive measures if and only if $\lambda _2 = - \frac{\lambda _1}{1+\lambda _1}$. Hence, this proposition follows from Propositions 19 and  14. $\square$

Proposition 21 can be stated in terms of the $\nu$-additive measure as follows.

Proposition 22

Let $Q_{\nu _1}$ and $Q_{\nu _2}$ be two $\nu$-additive measures on the finite set X. Then, $Q_{\nu _1}$ and $Q_{\nu _2}$ are a dual pair of belief- and plausibility measures on X if and only if

$$\begin{aligned} \nu _1+\nu _2 = 1. \end{aligned}$$

Proof

Based on Proposition 6, this proposition immediately follows from Proposition 21. $\square$

It should be added here that a $\nu$-additive measure may be supermodular or submodular depending on the value of its parameter $\nu$.

Definition 13

The set function $f: \mathscr {P}(X) \rightarrow \mathbb {R}$ on the finite set X is said to be submodular if

$$\begin{aligned} f(A)+f(B) \ge f(A \cup B) + f(A \cap B) \end{aligned}$$

holds for any $A,B \in \mathscr {P}(X)$.

Definition 14

The set function $f: \mathscr {P}(X) \rightarrow \mathbb {R}$ on the finite set X is said to be supermodular if

$$\begin{aligned} f(A)+f(B) \le f(A \cup B) + f(A \cap B) \end{aligned}$$

holds for any $A,B \in \mathscr {P}(X)$.

Corollary 6

A $\nu$-additive measure is supermodular if $\nu \in (0,1/2]$, and it is submodular if $\nu \in [1/2,1)$.

Proof

Since every belief measure is supermodular and every plausibility measure is submodular, this corollary immediately follows from Proposition 18. $\square$

5 A transformation between a $\nu$-additive ($\lambda$-additive) measure and a probability measure

Here, we will demonstrate that the $\nu$-additive ($\lambda$-additive) measures can be utilized for generating probability measures, and conversely, $\nu$-additive ($\lambda$-additive) measures can be generated from probability measures.

Definition 15

Let $\varSigma$ be a $\sigma$-algebra over the set X. Then, the function $\mu : \varSigma \rightarrow [0,\infty )$ is a measure on the space $(X, \varSigma )$ iff $\mu$ satisfies the following requirements:

• (1) $\forall A \in \varSigma : \mu (A) \ge 0$

• (2) $\mu (\emptyset ) = 0$

• (3) $\forall A_{1}, A_{2}, \ldots \in \varSigma$, if $A_i \cap A_j = \emptyset , \forall i\ne j$, then

  $$\begin{aligned} \mu \left( \bigcup \limits _{i=1}^{\infty } A_i \right) =\sum \limits _{i=1}^{\infty }\mu (A_i). \end{aligned}$$

Proposition 23

If $\varSigma$ is a $\sigma$-algebra over the set X, $Q_{\lambda }$ is a $\lambda$-additive measure, which satisfies the continuity property of monotone measures, on the space $(X, \varSigma )$, $\lambda >-\,1$, $\lambda \ne 0$, $c>0$ and the function $\hat{Q}_{\lambda ,c}: \varSigma \rightarrow [0, \infty )$ is given by

$$\begin{aligned} \hat{Q}_{\lambda ,c}(A) = c \ln (1+\lambda Q_{\lambda }(A)) \end{aligned}$$

for any $A \in \varSigma$, then $\hat{Q}_{\lambda ,c}$ is a measure on the space $(X, \varSigma )$.

Proof

$\hat{Q}_{\lambda ,c}(A)$ is trivially nonnegative for any $A \in \varSigma$ and if $A=\emptyset$, then $\hat{Q}_{\lambda ,c}(A) = 0$. That is, $\hat{Q}_{\lambda ,c}$ satisfies requirements (1) and (2) of Definition 15. Next, let $A_{1}, A_{2}, \ldots \in \varSigma$ be a countable collection of pairwise disjoint sets. Now, utilizing the definition of $\hat{Q}_{\lambda ,c}$, the fact that $Q_{\lambda }$ is a $\lambda$-additive measure on $(X, \varSigma )$ and Eq. (18), we get

$$\begin{aligned} \begin{aligned}&\hat{Q}_{\lambda ,c}\left( \bigcup \limits _{i=1}^{\infty } A_i \right) = c \ln \left( 1+\lambda \left( Q_{\lambda } \left( \bigcup \limits _{i=1}^{\infty } A_i \right) \right) \right) \\&\quad = c \ln \left( 1+\ \lambda \frac{1}{\lambda } \left( \prod \limits _{i=1}^{\infty } \left( 1+\lambda Q_{\lambda }(A_{i}) \right) -\,1 \right) \right) \\&\quad = \sum \limits _{i=0}^{\infty } c \ln \left( 1 + \lambda Q(A_{i})\right) = \sum \limits _{i=0}^{\infty } \hat{Q}_{\lambda ,c}(A_i). \end{aligned} \end{aligned}$$

It means that the function $\hat{Q}_{\lambda ,c}$ satisfies requirement (3) in Definition 15 as well. $\square$

Proposition 24

Let $\varSigma$ be a $\sigma$-algebra over the set X and let $Q_{\lambda }$ and $P_{\lambda }$ be two continuous functions on the space $(X, \varSigma )$ such that

$$\begin{aligned} P_{\lambda }(A) = \frac{\ln (1+\lambda Q_{\lambda }(A))}{\ln (1+\lambda )} \end{aligned}$$

(48)

holds for any $A \in \varSigma$, $\lambda > -\,1$, $\lambda \ne 0$. Then, $P_{\lambda }$ is a probability measure on $(X,\varSigma )$ if and only if $Q_{\lambda }$ is a $\lambda$-additive measure on $(X,\varSigma )$.

Proof

Firstly, we will show that if Eq. (48) holds and $Q_{\lambda }$ is a $\lambda$-additive measure on $(X, \varSigma )$, then $P_{\lambda }$ is a probability measure on $(X, \varSigma )$. Since $\forall A \in \varSigma : P_{\lambda }(A) = \hat{Q}_{\lambda ,c}(A)$ with $c= 1/\ln (1+\lambda )$, based on Proposition 23, $P_{\lambda }$ is a measure. Moreover, as $Q_{\lambda }(X)=1$, $P_{\lambda }(X)=1$ holds as well, and so the function $P_{\lambda }$ satisfies all the requirements of a probability measure given by Definition 4.

Secondly, we will show that if Eq. (48) holds and $P_{\lambda }$ is a probability measure on $(X, \varSigma )$, then $Q_{\lambda }$ is a $\lambda$-additive measure on $(X, \varSigma )$. Let $P_{\lambda }$ be a probability measure on $(X, \varSigma )$. From Eq. (48), we have

$$\begin{aligned} Q_{\lambda }(A) = \frac{1}{\lambda } \left( (1+\lambda )^{P_{\lambda }(A)}-\,1 \right) \end{aligned}$$

(49)

for any $A \in \varSigma$. Since $P_{\lambda }$ is a probability measure on $(X, \varSigma )$, $P_{\lambda }(X)=1$ and so from Eq. (49), we get $Q_{\lambda }(X)=1$. That is, $Q_{\lambda }$ satisfies requirement (1) of the $\lambda$-additive measures given by Definition 9. Now, let $A,B \in \varSigma$ such that $A \cap B = \emptyset$. Then, as $P_{\lambda }$ is a probability measure on $(X, \varSigma )$, the equation

$$\begin{aligned} P_{\lambda }(A \cup B) = P_{\lambda }(A) + P_{\lambda }(B) \end{aligned}$$

(50)

holds. Utilizing Eqs. (49) and (50), $Q_{\lambda }(A \cup B)$ can be written as

$$\begin{aligned} \begin{aligned} Q_{\lambda }(A \cup B)&= \frac{1}{\lambda } \left( (1+\lambda )^{P_{\lambda }(A \cup B)}-\,1 \right) \\&\quad = \frac{1}{\lambda } \left( (1+\lambda )^{P_{\lambda }(A)+P_{\lambda }(B)}-\,1 \right) \\&\quad = \frac{1}{\lambda } \left( (1+\lambda )^{P_{\lambda }(A)}-\,1 \right) + \frac{1}{\lambda } \left( (1+\lambda )^{P_{\lambda }(B)}-\,1 \right) \\&\qquad + \lambda \frac{1}{\lambda } \left( (1+\lambda )^{P_{\lambda }(A)}-\,1 \right) \frac{1}{\lambda } \left( (1+\lambda )^{P_{\lambda }(B)}-\,1 \right) \\&\quad = Q_{\lambda }(A) + Q_{\lambda }(B) + \lambda Q_{\lambda }(A) Q_{\lambda }(B). \end{aligned} \end{aligned}$$

It means that $Q_{\lambda }$ satisfies requirement (2) of the $\lambda$-additive measures given in Definition 9 as well; that is, $Q_{\lambda }$ meets all the requirements of a $\lambda$-additive measure. $\square$

Remark 8

The measure $P_{\lambda }$ is independent of the base of the logarithm because for any $A \in \varSigma$

$$\begin{aligned} \begin{aligned} \frac{\log _a(1+ \lambda Q_{\lambda } \left( A \right) ) }{\log _a(1+ \lambda )} = \frac{\frac{\log _{s}(1+ \lambda Q_{\lambda } \left( A \right) ) }{\log _{s}(a)} }{\frac{\log _{s}(1+ \lambda ) }{\log _{s}(a)} } = \frac{\log _{s}(1+ \lambda Q_{\lambda } \left( A \right) ) }{\log _{s}(1+ \lambda )}, \end{aligned} \end{aligned}$$

where $a,s >0$, $a,s \ne 1$, $\lambda >-1$ and $\lambda \ne 0$. Also, if $s = 1+\lambda$, then $P_{\lambda }(A) = \log _{1+\lambda } (1+\lambda Q_{\lambda }(A))$.

Utilizing the definition of the $\nu$-additive measure, Proposition 24 can be stated as follows.

Proposition 25

Let $\varSigma$ be a $\sigma$-algebra over the set X and let $Q_{\nu }$ and $P_{\nu }$ be two continuous functions on the space $(X, \varSigma )$ such that

$$\begin{aligned} P_{\nu }(A) = \frac{1}{2}\frac{\ln \left( 1+ \left( \left( \frac{1 - \nu }{\nu } \right) ^{2} -1\right) Q_{\nu }(A) \right) }{\ln \left( \frac{1-\nu }{\nu }\right) } \end{aligned}$$

(51)

holds for any $A \in \varSigma$, $\nu \in (0,1)$, $\nu \ne 1/2$. Then, $P_{\nu }$ is a probability measure on $(X,\varSigma )$ if and only if $Q_{\nu }$ is a $\nu$-additive measure on $(X,\varSigma )$.

Proof

Taking into account Proposition 6, this corollary immediately follows from Proposition 24. $\square$

Based on the result of Proposition 25, the formula in Eq. (51) may be viewed as a transformation between probability measures and $\nu$-additive measures.

6 Connections of $\nu$-additive ($\lambda$-additive) measures with other areas

6.1 Connection with rough sets

It is a well-known fact that the belief- and plausibility measures are connected with the rough set theory (see Dubois and Prade 1990; Yao and Lingras 1998; Wu et al. 2002). Here, we will show how the $\nu$-additive ($\lambda$-additive) measures are connected with the rough set theory.

Definition 16

Let X be a finite set, and let $R \subseteq X \times X$ be a binary equivalence relation on X. The pair $(\underline{R}(A), \overline{R}(A))$ is said to be the rough set of $A \subseteq X$ in the approximation space (X, R) if

$$\begin{aligned} \begin{aligned} \underline{R}(A)&= \lbrace x \in X \vert [x]_{R} \subseteq A \rbrace \\ \overline{R}(A)&= \lbrace x \in X \vert [x]_{R} \cap A \ne \emptyset \rbrace , \end{aligned} \end{aligned}$$

where $[x]_{R}$ is the R-equivalence class containing x.

The concept of a rough set was introduced by Pawlak (1982). The rough set $(\underline{R}(A), \overline{R}(A))$ can be utilized to characterize the set A by the pair of lower and upper approximations $(\underline{R}(A), \overline{R}(A))$. The lower approximation $\underline{R}(A)$ is the union of all elementary sets that are subsets of A, and the upper approximation $\overline{R}(A)$ is the union of all elementary sets that have a non-empty intersection with A. Note that the definitions of $\underline{R}(A)$ and $\overline{R}(A)$ are equivalent to the following statement: an element of X necessarily belongs to A if all of its equivalent elements belong to A, while an element of X possibly belongs to A if at least one of its equivalent elements belongs to A (Wu et al. 2002). Let the functions $\underline{q}, \overline{q}: \mathscr {P}(X) \rightarrow [0,1]$ be given as follows:

$$\begin{aligned} \underline{q}(A) = \frac{\vert \underline{R}(A) \vert }{\vert X \vert }, \text { } \overline{q}(A) = \frac{\vert \overline{R}(A) \vert }{\vert X \vert } \end{aligned}$$

for any $A \subseteq X$. Skowron (1989, 1990) showed that the functions $\underline{q}$ and $\overline{q}$ are a dual pair of belief- and plausibility measures and the corresponding basic probability assignment is $m(A^{*}) = \vert A^{*} \vert / \vert X \vert$ for all $A^{*} \in X/R$, and 0 otherwise. Furthermore, Yao and Lingras (1998) demonstrated that if Pl and Bl are a dual pair of plausibility and belief functions on X and m is the basic probability assignment of Bl satisfying the conditions: (1) the set of focal elements of m is a partition of X, (2) $m(A^{*}) = \vert A^{*} \vert / \vert X \vert$ for every focal element $A^{*}$ of m, then there exists an equivalence relation R on the set X, such that the induced qualities of upper and lower approximations satisfy

$$\begin{aligned} \underline{q}(A) = Bl(A), \text { } \overline{q}(A) = Pl(A) \end{aligned}$$

for any $A \subseteq X$ (Wu et al. 2002).

Based on these results and on our proposition findings, we will establish some connections between rough sets and $\nu$-additive measures by using the following propositions.

Proposition 26

Let $Q_{\nu _1}$ and $Q_{\nu _2}$ be two $\nu$-additive measures on the finite set X, and let $R \subseteq X \times X$ be a binary equivalence relation on X. Furthermore, let $(\underline{R}(A), \overline{R}(A))$ be the rough set of $A \in \mathscr {P}(X)$ with respect to the approximation space (X, R), and let the functions $\underline{q}, \overline{q}: \mathscr {P}(X) \rightarrow [0,1]$ be given by

$$\begin{aligned} \underline{q}(A) = \frac{\vert \underline{R}(A) \vert }{\vert X \vert }, \text { } \overline{q}(A) = \frac{\vert \overline{R}(A) \vert }{\vert X \vert }, \end{aligned}$$

where $\underline{R}(A)$ and $\overline{R}(A)$ are the lower- and upper approximations of A, respectively, for any $A \in \mathscr {P}(X)$. Then, if the equations

$$\begin{aligned} Q_{\nu _1}(A) = \underline{q}(A), \text { } Q_{\nu _2}(A) = \overline{q}(A), \end{aligned}$$

(52)

hold for any $A \in \mathscr {P}(X)$, then $Q_{\nu _1}$ and $Q_{\nu _2}$ are a dual pair of $\nu$-additive measures on X with $\nu _1 \in (0,1/2]$, $\nu _2 \in [1/2,1)$.

Proof

Based on Skowron’s results in Skowron (1989) and Skowron (1990), if the conditions of this proposition are satisfied, then the functions $\underline{q}$ and $\overline{q}$ are a dual pair of belief- and plausibility measures on X. Hence, the conditions that

• (i) $Q_{\nu _1}(A) = \underline{q}(A)$, $Q_{\nu _2}(A) = \overline{q}(A)$ hold for any $A \in \mathscr {P}(X)$

• (ii) $Q_{\nu _1}$ and $Q_{\nu _2}$ are two $\nu$-additive measures on X

and the fact that $\underline{q}$ and $\overline{q}$ are a dual pair of belief- and plausibility measures on X together imply that $Q_{\nu _1}$ and $Q_{\nu _2}$ are also a dual pair of $\nu$-additive measures on X. Furthermore, as $\underline{q}$ is a belief measure and $\overline{q}$ is a plausibility measure, based on Proposition 18, $\nu _1 \in (0,1/2]$ and $\nu _2 \in [1/2,1)$ hold as well. $\square$

Proposition 27

If $Q_{\nu _1}$ and $Q_{\nu _2}$ are a dual pair of $\nu$-additive measures on the finite set X with $\nu _1 \in (0,1/2]$, $\nu _2 \in [1/2,1)$ and m is a basic probability assignment that satisfies the conditions:

• (1) The set of focal elements of m is a partition of X

• (2) $m(A^{*}) = \vert A^{*} \vert / \vert X \vert$ for every focal element $A^{*}$ of m

• (3) $m(A^{*}) = \sum \limits _{B \subseteq A^{*}} (-1)^{\vert A^{*} \setminus B \vert } Q_{\nu _1}(B)$ for any $A^{*} \in \mathscr {P}(X)$,

then there exists an equivalence relation R on the set X, such that the equations

$$\begin{aligned} Q_{\nu _1}(A) = \underline{q}(A), \text { } Q_{\nu _2}(A) = \overline{q}(A) \end{aligned}$$

hold for any $A \in \mathscr {P}(X)$, where $(\underline{R}(A), \overline{R}(A))$ is the rough set of A with respect to the approximation space (X, R), $\underline{q}, \overline{q}: \mathscr {P}(X) \rightarrow [0,1]$ are given as

$$\begin{aligned} \underline{q}(A) = \frac{\vert \underline{R}(A) \vert }{\vert X \vert }, \text { } \overline{q}(A) = \frac{\vert \overline{R}(A) \vert }{\vert X \vert }, \end{aligned}$$

and $\underline{R}(A)$ and $\overline{R}(A)$ are the lower- and upper approximations of A, respectively.

Proof

Based on the result of Yao and Lingras (1998), if Pl and Bl are a dual pair of plausibility and belief functions on X and m is the basic probability assignment of Bl satisfying the conditions: (i) the set of focal elements of m is a partition of X, (ii) $m(A^{*}) = \vert A^{*} \vert / \vert X \vert$ for every focal element $A^{*}$ of m, then there exists an equivalence relation R on the set X, such that the induced qualities of upper and lower approximations satisfy

$$\begin{aligned} \underline{q}(A) = Bl(A), \text { } \overline{q}(A) = Pl(A) \end{aligned}$$

for any $A \in \mathscr {P}(X)$. Therefore, it is sufficient to show that if the conditions of our proposition are satisfied, then $Q_{\nu _1}$ is a belief measure on X, $Q_{\nu _2}$ is a plausibility measure on X, and m is the basic probability assignment of the belief measure $Q_{\nu _1}$.

Let us assume that the conditions of this proposition are satisfied. Then, since $Q_{\nu _1}$ and $Q_{\nu _2}$ are a dual pair of $\nu$-additive measures on the finite set X, based on Proposition 20, $Q_{\nu _1}$ and $Q_{\nu _2}$ are a dual pair of belief- and plausibility measures on X. Furthermore, as $\nu _1 \in (0,1/2]$ and $\nu _2 \in [1/2,1)$, based on Proposition 18, $Q_{\nu _1}$ is a belief measure on X and $Q_{\nu _2}$ is a plausibility measure on X, and so condition (3) means that m is the basic probability assignment of the belief measure $Q_{\nu _1}$. That is, we have shown that if the conditions of this proposition are satisfied, then all the conditions that are required to apply the result of Yao and Lingras (1998) are satisfied as well. $\square$

6.2 The $\lambda$-additive measure and the multi-attribute utility function

Here we will state interesting analogies between the $\lambda$-additive measure and the multi-attribute utility function. Let $X_1, X_2, \ldots , X_n$ be attributes, where each $X_i$ may be either a scalar attribute or a vector of scalar attributes $(i=1,2, \ldots , n)$. Furthermore, let the consequence space X be a rectangular subset of the n-dimensional Euclidean space. Then, a specific consequence may be given by a vector $(x_1, x_2, \ldots , x_n)$, where $x_i$ is a particular value of the attribute $X_i$$(i=1,2, \ldots , n)$. The utility function $u: X \rightarrow \mathbb {R}$, which is assumed to be continuous, assigns a utility value to the consequence $(x_1, x_2, \ldots , x_n)$; that is, the utility of consequence $(x_1, x_2, \ldots , x_n)$ is $u(x_1, x_2, \ldots , x_n)$ (Keeney 1974). Here, we will utilize the concept of the utility independence of attributes (see, e.g., Keeney and Raiffa 1993).

Definition 17

Attribute $X_i$ is utility independent of attribute $X_j$ if conditional preferences for lotteries over $X_i$ given a fixed value for $X_j$ do not depend on the particular value of $X_j$.

Keeney and Raiffa (1993) proved the following proposition which states that the mutual utility independence of attributes implies a multiplicative multi-attribute utility function.

Proposition 28

If $X_1, X_2, \ldots , X_n$ are mutually utility independent attributes, then

$$\begin{aligned} u_M(x_1,x_2, \ldots ,x_n) = \frac{1}{k} \left( \prod \limits _{i=1}^{n} \left( 1+kk_iu_i(x_i)\right) -1 \right) , \end{aligned}$$

(53)

where $u_M: \mathbb {R}^{n} \rightarrow [0,1]$ is a multi-attribute utility function, $u_i: \mathbb {R} \rightarrow [0,1]$ are utility functions, $k_i$ is the weight of attribute $X_i$ with $0< k_i <1$, and $k>-1$, $k \ne 0$ is a scaling constant $(i=1,2, \ldots , n)$.

Proof

See Keeney and Raiffa (1993). $\square$

The multi-attribute utility function $u_M$ in Eq. (53) plays a key role in multi-attribute utility theory and can be written as

$$\begin{aligned} 1+ku_M(x_1,x_2, \ldots ,x_n) = \prod _{i=1}^{n} \left( 1+kk_iu_i(x_i)\right) . \end{aligned}$$

(54)

If k is positive in Eq. (54), then $u^{*}(x_1, x_2, \ldots , x_n)= 1+k u_M(x_1, x_2, \ldots , x_n)$ is a multi-attribute utility function, $u_i^{*}(x_i)=1+kk_iu_i(x_i)$ are utility functions and $u^{*}(x_1, x_2, \ldots , x_n) = \prod _{i=1}^{n} u_i^{*}(x_i)$, where $i=1,2, \ldots , n$. Similarly, if k is negative in Eq. (54), then $u^{*}(x_1, x_2, \ldots , x_n)= -(1+k u_M(x_1, x_2, \ldots , x_n))$ is a multi-attribute utility function, $u_i^{*}(x_i)=-(1+kk_iu_i(x_i))$ are utility functions and $-u^{*}(x_1, x_2, \ldots , x_n) = (-1)^{n} \prod _{i=1}^{n} u_i^{*}(x_i)$, where $i=1,2, \ldots , n$. That is, Eq. (54) describes a multiplicative relationship between the multi-attribute utility function and the individual univariate utility functions. Hence, Eq. (53) is referred to as the multi-attribute multiplicative utility function.

We can see that the right-hand side of Eq. (18) with $\lambda >-1$, $\lambda \ne 0$ has the same form as the right-hand side of Eq. (53). It means that there is an interesting connection between the $\lambda$-additive measures and the multi-attribute multiplicative utility function. Namely, a $\lambda$-additive measure with $\lambda \ne 0$ of the union of n pairwise disjoint sets is computed in the same way as the multi-attribute utility of n mutually utility independent attributes.

Here, the formula in Eq. (53) can be written as

$$\begin{aligned} \begin{aligned}&u_M(x_1,x_2, \ldots ,x_n) = \sum \limits _{i=1}^{n} k_i u_i(x_i) \\&\quad + \sum \limits _{r=2}^{n} k^{r-1} \sum \limits _{\begin{array}{c} 1 \le i_{1}< \cdots \\ \cdots <i_{r} \le n \end{array}} k_{i_1} \cdots k_{i_r} u_{i_1}(x_{i_1}) \cdots u_{i_r}(x_{i_r}) \end{aligned} \end{aligned}$$

(55)

from which

$$\begin{aligned} \lim \limits _{k \rightarrow 0} u_M(x_1,x_2, \ldots ,x_n) = \sum \limits _{i=1}^{n} k_i u_i(x_i). \end{aligned}$$

Note that

$$\begin{aligned} u_A(x_1,x_2, \ldots ,x_n) = \sum \limits _{i=1}^{n} k_i u_i(x_i) \end{aligned}$$

(56)

is the so-called multi-attribute additive utility function (Keeney 1974). We can get Eq. (56) from Eq. (55) by allowing for $k=0$.

Definition 18

Two attributes $X_i$ and $X_j$ are additive independent if the paired preference comparison of any two lotteries, defined by two joint probability distributions on $X_i \times X_j$, depends only on their marginal distributions.

It can be shown that if and only if the preferences over lotteries on attributes $X_1, X_2, \ldots , X_n$ depend only on their marginal probability distributions (i.e., the attributes are additive independent), then the n-attribute utility function is additive (Keeney and Raiffa 1993).

Notice that the right-hand side of Eq. (18) with $\lambda =0$ has the same form as the right-hand side of Eq. (56). It means that a $\lambda$-additive measure with $\lambda = 0$ of the union of n pairwise disjoint sets is computed in the same way as the multi-attribute utility of n additive independent attributes.

Table 1 summarizes the analogies between the $\lambda$-additive measures and the multi-attribute utility functions.

Table 1 $\lambda$-Additive measure of union of pairwise disjoint sets and utility value of consequence $(x_1, x_2, \ldots , x_n)$

6.3 The $\lambda$-additive measure and some operators of continuous-valued logic

Here, we will state a formal connection between the $\lambda$-additive measure and certain operators of continuous-valued logic.

Definition 19

The generalized Dombi operator $o_{GD,\gamma }^{(\alpha )}: [0,1]^{n} \rightarrow [0,1]$ is given by

$$\begin{aligned} \begin{aligned}&o_{GD,\gamma }^{(\alpha )}(\mathbf {x}) \\&\quad = \frac{1}{1+ \left( \frac{1}{\gamma } \left( \prod _{i=1}^{n} \left( 1+ \gamma \left( \frac{1-x_i}{x_i} \right) ^{\alpha } \right) -1 \right) \right) ^{1/\alpha }}, \end{aligned} \end{aligned}$$

(57)

where $\mathbf {x} = (x_1,x_2, \ldots , x_n)$, $x_1,x_2, \ldots , x_n$ are continuous-valued logic variables, $\alpha \in (-\infty , \infty )$ and $\gamma \in (0,\infty )$ (Dombi 2008).

It can be shown that if $\alpha >0$, then $o_{GD,\gamma }^{(\alpha )}$ is a conjunction operator, and if $\alpha <0$, then $o_{GD,\gamma }^{(\alpha )}$ is a disjunction operator (see Dombi 2008). Moreover, the operator $o_{GD,\gamma }^{(\alpha )}$ is general because depending on its parameter values it can cover a range of familiar fuzzy conjunction and disjunction operators including the Dombi operators (Dombi 1982), the product operators (Dombi 2008), the Einstein operators (Wang and Liu 2012), the Hamacher operators (Hamacher 1978), the drastic operators (Zimmermann 2013) and the min–max operators (Zadeh 1965). Table 2 summarizes the operators that the generalized Dombi operator class can cover.

Table 2 Operators covered by the generalized Dombi operator class

Here, from Eq. (57) we have

$$\begin{aligned} \begin{aligned}&\left( \frac{1-o_{GD,\gamma }^{(\alpha )}(\mathbf {x})}{o_{GD,\gamma }^{(\alpha )}(\mathbf {x})} \right) ^{\alpha } \\&\quad = \frac{1}{\gamma } \left( \prod _{i=1}^{n} \left( 1+ \gamma \left( \frac{1-x_i}{x_i} \right) ^{\alpha } \right) -1 \right) \end{aligned} \end{aligned}$$

(58)

for any $o_{GD,\gamma }^{(\alpha )}(\mathbf {x}) \in (0,1]$. Next, the generator function $g: (0,1] \rightarrow [0,\infty )$ of Dombi operators (Dombi 1982) is given by

$$\begin{aligned} g(x) = \left( \frac{1-x}{x} \right) ^{\alpha }. \end{aligned}$$

Utilizing this function, Eq. (58) can be written as

$$\begin{aligned} \begin{aligned} g\left( o_{GD,\gamma }^{(\alpha )}(x_1, x_2, \ldots , x_n) \right) = \frac{1}{\gamma } \left( \prod _{i=1}^{n} \left( 1+ \gamma g(x_i) \right) -1 \right) . \end{aligned} \end{aligned}$$

(59)

Recall that based on Proposition 8, if X is a finite set, $Q_{\lambda }$ is a $\lambda$-additive measure on X, $\lambda >-1$, $\lambda \ne 0$ and $A_1, A_2, \ldots , A_n \in \mathscr {P}(X)$ are pairwise disjoint sets, then

$$\begin{aligned} Q_{\lambda } \left( \bigcup \limits _{i=1}^n A_i \right) = \frac{1}{\lambda } \left( \prod \limits _{i=1}^{n} (1+\lambda Q_{\lambda }( A_i))-1 \right) . \end{aligned}$$

(60)

From Eqs. (59) and (60), we notice an interesting analogy. Namely, a $\lambda$-additive measure with $\lambda \ne 0$ of the union of n pairwise disjoint sets is computed in the same way as the value of the generator function of Dombi operator for the value of the generalized Dombi operation over n continuous-valued logic variables. It should be added that this analogy is just a formal one since $g(x_i) \in (0, \infty )$ and $Q_{\lambda }(A_i) \in [0,1]$, and $g(x_i)$ and $Q_{\lambda }(A_i)$ have different meanings.

7 Summary and future plans

In our study, we introduced the $\nu$-additive measure as an alternatively parameterized $\lambda$-additive measure. Here, we will summarize our main findings concerning the $\nu$-additive ($\lambda$-additive) measures.

• (1) A $\nu$-additive measure and a $\lambda$-additive measure (Sugeno $\lambda$-measure) are identical if and only if

  $$\begin{aligned} \lambda = \left( \frac{1 - \nu }{\nu } \right) ^{2} -1, \end{aligned}$$

  where $\lambda \in (-1,\infty )$, $\nu \in (0,1)$.

• (2) Two $\nu$-additive measures are a dual pair if and only if the sum of their parameters equals 1.

• (3) A $\nu$-additive measure is a

  • (a) belief measure if and only if $0<\nu \le 1/2$

  • (b) probability measure if and only if $\nu =1/2$

  • (c) plausibility measure if and only if $1/2 \le \nu <1$.

• (4) Two $\nu$-additive measures are a dual pair of belief- and plausibility measures if and only if the sum of their parameters equals 1.

• (5) There exists a transformation that can be utilized for transforming a $\nu$-additive ($\lambda$-additive) measure into a probability measure; conversely, this transformation can be utilized for transforming a probability measure into a $\nu$-additive ($\lambda$-additive) measure.

• (6) Dual pairs of $\nu$-additive measures are strongly associated with the lower- and upper approximation pairs of rough sets.

• (7) There are interesting formal connections between the $\lambda$-additive measures and the multi-attribute utility functions. Namely,

  • (a) if $\lambda = 0$, then the $\lambda$-additive measure of the union of n pairwise disjoint sets is computed in the same way as the multi-attribute utility of n additive independent attributes

  • (b) if $\lambda >-1$ and $\lambda \ne 0$, then the $\lambda$-additive measure of the union of n pairwise disjoint sets is computed in the same way as the multi-attribute utility of n mutually utility independent attributes.

• (8) There is an interesting formal connection between the $\lambda$-additive measure and certain operators of continuous-valued logic. Namely, if $\lambda >-1$ and $\lambda \ne 0$, then the computation method of $\lambda$-additive measure of union of n pairwise disjoint sets is identical with that of the generator function of the Dombi operator at the value of the generalized Dombi operation over n continuous-valued logic variables.

As part of our future research plans, we would like to formulate a calculus of the $\nu$-additive measure and generalize the Bayes theorem and the Poincaré formula for $\nu$-additive measures. We also plan to study how the $\nu$-additive measure can be utilized in the fields of computer science, engineering and economics. In particular, we aim to investigate the potential application of $\nu$-additive measures in network science.