Existence of competitive equilibrium in coalition production economies with a continuum of agents

Abstract

Aumann (Econometrica 34: 1–17, 1966) established the existence of competitive equilibria in exchange economies with a continuum of traders. This result has been extended to special production economies under the condition that the production sets are countably additive or superadditive in the literature. In this paper, we extend Aumann’s result to coalition production economies with a continuum of traders in which each coalition can have a different production set and the production sets need not to be additive or superadditive. Our new existence theorem for competitive equilibrium implies the classical result by Hildenbrand (Econometrica 38: 608–623, 1970), and implies those by Sondermann (J Econ Theory 8: 259–291, 1974) and Greenberg et al. (J Math Econ 6: 31–41, 1979) under the assumption that the production sets are convex.

Appendix I

In this appendix, we will give a proof for Auxiliary Theorem. We need the following 3-person game $\Gamma$ [see Greenberg et al. (1979)] and Debreu’s lemma from Debreu (1952).

A 3-person game $\Gamma$ consists of

• (1) three sets $K_{1}$, $K_{2}$, and $K_{3}$;

• (2) three correspondences ${\mathcal {A}}_{1}: K_{2}\times K_{3} \mapsto K_{1}$, ${\mathcal {A}}_{2}: K_{1}\times K_{3} \mapsto K_{2}$, ${\mathcal {A}}_{3}: K_{1}\times K_{2} \mapsto K_{3}$;

• (3) three functions $u_{i}: K \rightarrow {\mathbb {R}}$, $i = 1, 2, 3$, where $K = K_{1}\times K_{2} \times K_{3}$.

An equilibrium for $\Gamma$ is a $\overline{k} = (\overline{k}_{1}, \overline{k}_{2}, \overline{k}_{3}) \in K$ such that $\overline{k}_{1} \in {\mathcal {A}}_{1}(\overline{k}_{2}, \overline{k}_{3})$, $\overline{k}_{2} \in {\mathcal {A}}_{2}(\overline{k}_{1}, \overline{k}_{3})$, $\overline{k}_{3} \in {\mathcal {A}}_{3}(\overline{k}_{1}, \overline{k}_{2})$, and $u_{1}(\overline{k}) \ge u_{1}(k_{1}, \overline{k}_{2}, \overline{k}_{3})$, $u_{2}(\overline{k}) \ge u_{2}(\overline{k}_{1}, k_{2}, \overline{k}_{3})$, $u_{3}(\overline{k}) \ge u_{3}(\overline{k}_{1}, \overline{k}_{2}, k_{3})$ for every $k_{1} \in {\mathcal {A}}_{1}(\overline{k}_{2}, \overline{k}_{3})$, $k_{2} \in {\mathcal {A}}_{2}(\overline{k}_{1}, \overline{k}_{3})$, and $k_{3} \in {\mathcal {A}}_{3}(\overline{k}_{1}, \overline{k}_{2})$.

Lemma A.1

(Debreu 1952) Let $\Gamma$ be a 3-person game which satisfies, for $i = 1, 2, 3$,

• (D.1) $K_{i}$ is nonempty convex compact subset of a Euclidean space;

• (D.2) ${\mathcal {A}}_{i}$ is a continuous correspondence, convex and nonempty valued;

• (D.3) $u_{i}$ is continuous and quasi-concave in the ith coordinate.

Then $\Gamma$ has an equilibrium.

The following proof is a modification of the existence proof by Greenberg et al. (1979), the main difference is at the part of proving $\int _{T}(\mathbf{f}(t) - \mathbf{w}(t))dt = y$ (while in the proof by Greenberg et al., it only needs $\int _{T}(\mathbf{f}(t) - \mathbf{w}(t))dt \le \int _{T}g(t)dt = y$, we must have $\int _{T}(\mathbf{f}(t) - \mathbf{w}(t))dt = y$ as required by a competitive equilibrium which ensures that $\int _{T}(\mathbf{f}(t) - \mathbf{w}(t))dt$ maximizes the total profit on Y at a competitive equilibrium $(p, \mathbf{f}, y)$). We also need to pay special attention to the fact that an allocation here must satisfy (2.1) which is not required by Greenberg et al. (1979). Let V and W be topological spaces, and let $F : V \mapsto 2^{W}$ be a measurable correspondence. F is uniformly compact near a point $x \in V$ if x has an open neighborhood A such that $cl_{W}(\cup _{y \in A}F(y))$ is compact, where $cl_{W}(B)$ is the closure of B in W. F is uniformly compact if F is uniformly compact near x for all $x \in V$.

Recall that

$$\begin{aligned} P = \left\{ p \in {\mathbb {R}}_{+}^{l}: \sum _{i = 1}^{l}p^{i} = 1\right\} . \end{aligned}$$

Set

$$\begin{aligned}&P^{*} = \bigg \{p \in P: \sup p \cdot Y < \infty , \text{ there } \text{ exists } \text{ a } \text{ production } y \in Y \text{ such } \text{ that } p \cdot y = \sup p \cdot Y\\&\text{ and } \text{ for } \text{ any } S \in {\mathcal {F}}, \sup p \cdot Y^{S} \le (\sup p \cdot Y) \int _{S} \beta (t,p) dt\bigg \}. \end{aligned}$$

Here $Y^{S}$ and Y should be understood as $Y^{S}_{m}$ and $Y_{m}$ in the truncated coalition production economy ${\mathcal {E}}^{m}$.

Remark E

With the condition that Y is closed and $\beta (t,p)$ is continuous with respect to p, one can easily show that the set $P^{*}$ is closed and thus compact. In fact, let $p_{m} \in P^{*}$ for $m \ge 1$ be such that $p_{m} \rightarrow p$. Then it is easy to prove that $p \in P^{*}$.

As noted by a referee, for any $p \in P$, there exists $y \in Y_{m}$ such that $p \cdot y = \sup p \cdot Y_{m} = \max p \cdot Y_{m}$ for all $m \ge 1$ since $Y_{m}$ is compact. Here is the well-known supporting hyperplane theorem.

Supporting Hyperplane Theorem. Let $D \subseteq {\mathbb {R}}^{l}$ be convex and let z be a boundary point of D. Then there exists a supporting hyperplane $H = \{x \in {\mathbb {R}}^{l}:a \cdot x = k\}$ containing z (with $0 \ne a = (a^{1}, a^{2}, \ldots , a^{l}) \in {\mathbb {R}}^{l}$ and k being a constant), that is, $a \cdot x \le a \cdot z$ for every $x \in D$.

Proposition A

Assumptions (P.1)–(P.3) imply that $P^{*} \ne \emptyset$.

Proof

By assumptions (P.1), $Y = Y^{T}$ is a convex set satisfying $Y \cap {\mathbb {R}}_{+}^{l} = \{0\}$ which implies that the origin 0 is a boundary point of Y. Applying the supporting hyperplane theorem with $z = 0$ and $D = Y$, there exists a nonzero $a = (a^{1}, a^{2}, \dots , a^{l}) \in {\mathbb {R}}^{l}$ such that $a \cdot y \le 0$ for all $y \in Y$. By the free disposal condition in (P.2), we have ${\mathbb {R}}_{-}^{l} \subseteq Y$ which implies that $a^{i} \ge 0$ for every $1 \le i \le l$. Set $p = \frac{a}{||a||}$, where ||a|| denotes the length of a. Then $p \in P$ and $p \cdot y \le 0$ for all $y \in Y$. Thus, $\sup p \cdot Y = 0$. By assumptions (P.1) and (P.3), we have $0 \in Y^{S} \subseteq Y$ which implies that $\sup p \cdot Y^{S} = 0$ for every coalition $S \in {\mathcal {F}}$. Thus $p \in P^{*}$. $\Box$

Proof of Auxiliary Theorem. For convenience, let us still use X to denote $X^{m}$ and $Y^{S}$ to denote $Y^{S}_{m}$ for each coalition $S \in {\mathcal {F}}$. Then under the assumptions (II.1)–(II.4), (P.1)–(P.3), we have that X is uniformly compact, X(t) is convex for almost every t in T, Y is uniformly compact, $0 \in Y^{S}$ for all $S \in {\mathcal {F}}$, $\mathbf{w}(t) \in int(X(t))$ for almost every $t \in T$, and $U_{t}$ is continuous for almost every $t \in T$, where $U_{t}: X \times P \rightarrow {\mathbb {R}}$ is a price dependent continuous and quasi-concave function which represents the preference relation $\succ _{t}$ [real valued continuous and quasi-concave utility functions can be used to approximate rather general preference relations arbitrarily closely according to Billera (1974), see also Arrow and Debreu (1954)]. Without loss of generality, we can assume that all values of every $U_{t}$ are contained in the unit interval [0, 1].

Define the 3-person game $\Gamma$ as follows: $K_{1} = P^{*}$ (which is compact by Remark E), $K_{2} = \int X dt \times [0, 1]$, $K_{3} = Y$. For $p \in K_{1}$, $(x, \alpha ) \in K_{2}$, and $y \in K_{3}$, let

$$\begin{aligned} {\mathcal {A}}_{1}((x, \alpha ), y)= & {} K_{1},\\ {\mathcal {A}}_{2}(p, y)= & {} \left\{ (x, \alpha ) \in K_{2} : \begin{array}{llll} \text{ there } \text{ exists } \text{ an } \text{ allocation } \mathbf{f} \text{ such } \text{ that } \\ x = \int _{T} \mathbf{f}(t) dt, \alpha = \int _{T}U_{t}(\mathbf{f}(t), p)dt,\\ p \cdot \mathbf{f}(t) \le p \cdot \mathbf{w}(t) + (\sup p \cdot Y)\beta (t,p) \text{ for } \text{ a.e. } t \in T,\\ \sup p \cdot Y^{S} \le (\sup p \cdot Y)\int _{S}\beta (t,p)dt \text{ for } \text{ every } \text{ coalition } S \in {\mathcal {F}} \end{array} \right\} ,\\ {\mathcal {A}}_{3}(p, (x, \alpha ))= & {} K_{3}, \end{aligned}$$

and

$$\begin{aligned} u_{1}(p, (x, \alpha ), y)= & {} p \cdot \left( x - \int _{T}{} \mathbf{w}(t)dt - y\right) ,\\ u_{2}(p, (x, \alpha ), y)= & {} \alpha , \\ u_{3}(p, (x, \alpha ), y)= & {} p \cdot y. \end{aligned}$$

It can be verified in almost the same way as in the proof by Greenberg et al. (1979) that $\Gamma$ satisfies (D.1)-(D.3) as long as the following are noticed and taken care of: (1) replace $Max p \cdot Y(t)$ and $Max p_{n} \cdot Y(t)$ by $(\sup p \cdot Y)\beta (t,p)$ and $(\sup p_{n} \cdot Y)\beta (t,p)$, respectively; (2) pay an extra effort to handle the additional condition $\sup p \cdot Y^{S} \le (\sup p \cdot Y)\int _{S}\beta (t,p)dt$ for every $S \in {\mathcal {F}}$ in ${\mathcal {A}}_{2}(p, y)$ wherever needed; (3) the nonemptiness of ${\mathcal {A}}_{2}$ is guaranteed by Proposition A and the fact that $(\int _{T}{} \mathbf{w}(t) dt, \int _{T}U_{t}(\mathbf{w}(t), p)dt) \in {\mathcal {A}}_{2}(p, y)$; (4) if $\{\mathbf{f}_{n}\}$ is a sequence of allocations and $\mathbf{f}_{n} \rightarrow \mathbf{f}$, then $\mathbf{f}$ is also an allocation: by (2.1), $\mathbf{f}_{n}$ are allocations imply that $\mathbf{f}_{n} \in X$ and $\int _{T}{} \mathbf{f}_{n}(t)dt - \int _{T}{} \mathbf{w}(t)dt = y_{n} \in Y$. Since X is compact, by Fatou’s lemma in several dimensions [see Hildenbrand (1974), p. 69], the sequence $\{\mathbf{f}_{n}\}$ has a convergent subsequence, say $\mathbf{f}_{n} \rightarrow \mathbf{f} \in X$; since Y is compact and $y_{n} \in Y$ for all $n \ge 1$, $\{y_{n}\}$ has a convergent subsequence, say, $y_{n} \rightarrow y \in Y$. It follows that $\int _{T}{} \mathbf{f}(t)dt - \int _{T}{} \mathbf{w}(t)dt = y \in Y$ and so $\mathbf{f}$ is an allocation. For example, we will prove ${\mathcal {A}}_{2}$ is convex valued through Lemma A.2 below.

By Lemma A.1, $\Gamma$ has an equilibrium $(p, (x, \alpha ), y)$, that is, there exists an allocation $\mathbf{f}$ such that $x = \int _{T} \mathbf{f}(t) dt$, $\alpha = \int _{T}U_{t}(\mathbf{f}(t), p)dt$, $p \cdot \mathbf{f}(t) \le p \cdot \mathbf{w}(t) + (\sup p \cdot Y)\beta (t,p)$ for a.e. $t \in T$, and $\sup p \cdot Y^{S} \le (\sup p \cdot Y)\int _{S}\beta (t,p)dt \text{ for } \text{ every } S \in {\mathcal {F}}$. We now show that (p, f, y) is a competitive equilibrium for the economy ${\mathcal {E}}^{m}$.

By the definition of $K_{3}$ and $u_{3}$, $(p, (x, \alpha ), y)$ is an equilibrium of $\Gamma$ implies that

$$\begin{aligned} p \cdot y = \sup p \cdot Y, \end{aligned}$$

together with the condition that $\sup p \cdot Y^{S} \le (\sup p \cdot Y)\int _{S}\beta (t,p)dt \text{ for } \text{ every } S \in {\mathcal {F}}$, condition (ii) for a competitive equilibrium is satisfied.

Next we show that $\int _{T}(\mathbf{f}(t) - \mathbf{w}(t))dt = y$ as required by condition (i) for a competitive equilibrium. Set $b = \int _{T}(\mathbf{f}(t) - \mathbf{w}(t))dt - y = \int _{T}[\mathbf{f}(t) - \mathbf{w}(t) - y\beta (t,p)]dt$ (where $y = \int _{T}y\beta (t,p)dt$ since $\int _{T}\beta (t,p)dt = 1$). We will prove that $b = 0$. Since $p \cdot \mathbf{f}(t) \le p \cdot \mathbf{w}(t) + (\sup p \cdot Y)\beta (t,p)$ for a.e. $t \in T$, we have

$$\begin{aligned} p \cdot \mathbf{f}(t) \le p \cdot \mathbf{w}(t) + (p \cdot y)\beta (t,p) \quad \text{ for } \text{ a.e. } t \in T \end{aligned}$$

which implies that $p \cdot \int _{T}(\mathbf{f}(t) - \mathbf{w}(t))dt \le (p \cdot y)\int _{T}\beta (t,p)dt = p \cdot y$, as $\int _{T}\beta (t,p)dt = 1$. Thus, $p \cdot [x - \int _{T}{} \mathbf{w}(t)dt - y] = p \cdot [\int _{T}(\mathbf{f}(t) - \mathbf{w}(t))dt - y] \le 0$. By the definition of equilibrium in $\Gamma$, for every $p' \in P^{*}$,

$$\begin{aligned} p \cdot \left( x - \int _{T}{} \mathbf{w}(t)dt - y\right)= & {} u_{1}(p, (x, \alpha ), y) \ge u_{1}(p', (x, \alpha ), y)\\= & {} p' \cdot \left( x - \int _{T}{} \mathbf{w}(t)dt - y\right) . \end{aligned}$$

Thus $p' \cdot [\int _{T}(\mathbf{f}(t) - \mathbf{w}(t))dt - y] = p' \cdot (x - \int _{T}{} \mathbf{w}(t)dt - y) \le 0$ for every $p' \in P^{*}$. It follows that $b = \int _{T}(\mathbf{f}(t) - \mathbf{w}(t))dt - y = \int _{T}[\mathbf{f}(t) - \mathbf{w}(t) - y\beta (t,p)]dt \le 0$. If $b \ne 0$, then $b^{i} < 0$ for some $1 \le i \le l$. For each i with $b^{i} < 0$, let $S_{i} = \{t : \mathbf{f}^{i}(t) < \mathbf{w}^{i}(t) + y^{i}\beta (t,p)\}$. Then $\mu (S_{i}) > 0$ (for otherwise $b^{i} = \int _{T}[\mathbf{f}^{i}(t)dt - \mathbf{w}^{i}(t) - y^{i}\beta (t,p)]dt = 0$), where $\mu$ is Lebesgue measure. Clearly, $|b^{i}|= |\int _{T}[\mathbf{f}^{i}(t) - \mathbf{w}^{i}(t)]dt - y^{i}| = |\int _{T}[\mathbf{f}^{i}(t) - \mathbf{w}^{i}(t) - y^{i}\beta (t,p)]dt| = \int _{T}[\mathbf{w}^{i}(t) + y^{i}\beta (t,p) - \mathbf{f}^{i}(t)]dt \leqq \int _{S_{i}}[\mathbf{w}^{i}(t) + y^{i}\beta (t,p) - \mathbf{f}^{i}(t)]dt = c^{i}$. We construct $\mathbf{z}(t)$ as follows: for each i with $b^{i} < 0$, define

$$\begin{aligned} \mathbf{z}^{i}(t) = \left\{ \begin{array}{ll} \mathbf{f}^{i}(t)+ \frac{|b^{i}|}{c^{i}}(\mathbf{w}^{i}(t) - \mathbf{f}^{i}(t) + y^{i}\beta (t,p)) &{}\quad \text{ for } t \in S_{i},\\ \mathbf{f}^{i}(t), &{} \quad \text{ for } t \not \in S_{i}; \end{array}\right. \end{aligned}$$

and for each i with $b^{i} = 0$, let $\mathbf{z}^{i}(t) = \mathbf{f}^{i}(t)$. Since $p \cdot \mathbf{f}(t) \le p \cdot \mathbf{w}(t) + (\sup p \cdot Y)\beta (t,p) = p \cdot \mathbf{w}(t) + (p \cdot y)\beta (t,p)$ for a.e. $t \in T$ and $|b_{i}| \le c_{i}$, it follows that $p \cdot \mathbf{z}(t) \le p \cdot \mathbf{w}(t) + (p \cdot y)\beta (t,p) = p \cdot \mathbf{w}(t) + (\sup p \cdot Y)\beta (t,p)$ for a. e. $t \in T$. Moreover, since $|b^{i}|= |\int _{T}[\mathbf{f}^{i}(t) - \mathbf{w}^{i}(t)]dt - y^{i}| = \int _{T}{} \mathbf{w}^{i}(t)dt + y^{i} - \int _{T}{} \mathbf{f}^{i}(t)dt$ for each i with $b^{i} < 0$ and $\mathbf{z}^{i}(t) = \mathbf{f}^{i}(t)$ for each i with $b^{i} = 0$, we have

$$\begin{aligned} \int _{T}{} \mathbf{z}^{i}(t)dt = \int _{T}{} \mathbf{w}^{i}(t)dt + y^{i} \quad \text{ for } \text{ each } 1 \le i \le m \end{aligned}$$

which implies that $\mathbf{z}$ is an allocation satisfying

$$\begin{aligned} \int _{T}[\mathbf{z}(t) - \mathbf{w}(t)]dt = y. \end{aligned}$$

For each i with $b^{i} < 0$, we have $\mathbf{z}^{i}(t) > \mathbf{f}^{i}(t)$ for all $t \in S_{i}$ as $\mathbf{w}^{i}(t) - \mathbf{f}^{i}(t) + y^{i}\beta (t,p) > 0$. Because of desirability (II.2), we have $U_{t}(\mathbf{z}(t), p) > U_{t}(\mathbf{f}(t), p)$ for all $t \in S_{i}$. It follows that $\alpha ' = \int _{T}U_{t}(\mathbf{z}(t), p)dt > \int _{T}U_{t}(\mathbf{f}(t), p)dt = \alpha$. But $(p, (x, \alpha ), y)$ is an equilibrium of $\Gamma$ implies that $\alpha = u_{2}(p, (\int _{T}{} \mathbf{f}(t)dt, \int _{T}U_{t}(\mathbf{f}(t), p)dt), y) \ge u_{2}(p, (\int _{T}{} \mathbf{z}(t)dt, \int _{T}U_{t}(\mathbf{z}(t), p)dt), y) = \alpha '$, a contradiction. Thus, we must have $b = 0$, that is, $\int _{T}(\mathbf{f}(t) - \mathbf{w}(t))dt = y$, and so $\mathbf{f}$ is an allocation such that $\int _{T}(\mathbf{f}(t) - \mathbf{w}(t))dt$ maximizes the total profit on Y as $p \cdot y = \sup p \cdot Y$.

To complete the proof, we need to show that condition (iii) for a competitive equilibrium holds. Recall that $p \cdot \mathbf{f}(t) \le p \cdot \mathbf{w}(t) + (\sup p \cdot Y)\beta (t,p)$ for a.e. $t \in T$. We only need to prove that $U_{t}(\mathbf{a}, p) > U_{t}(\mathbf{f}(t), p)$ implies that $p \cdot \mathbf{a} > p \cdot \mathbf{w}(t) + (\sup p \cdot Y)\beta (t,p)$, which can be proved similarly as in the existence proof by Greenberg et al. (1979). $\Box$

Liapunov Theorem. Let $(T, {\mathcal {F}}, \mu )$ be an atomless finite measure space and $\mathbf{f}$ is an integrable function from T into ${\mathbb {R}}^{l}$. Then the set $\{\int _{S}{} \mathbf{f}d\mu : S \in {\mathcal {F}}\}$ is a convex set in ${\mathbb {R}}^{l}$.

Lemma A.2

${\mathcal {A}}_{2}$ is convex valued.

Proof

Let $p \in K_{1}$, $y \in K_{3}$, $(x, \alpha ) \in {\mathcal {A}}_{2}(p, y)$, $(x', \alpha ') \in {\mathcal {A}}_{2}(p, y)$, and let $0 \le \lambda \le 1$. That is, there exists allocations $\mathbf{f}(t)$ and $\mathbf{f}'(t)$ such that $x = \int _{T}{} \mathbf{f}(t)dt$, $x' = \int _{T}{} \mathbf{f}'(t)dt$, $\alpha = \int _{T}U_{t}(\mathbf{f}(t), p)dt$, $\alpha ' = \int _{T}U_{t}(\mathbf{f}'(t), p)dt$, $p \cdot \mathbf{f}(t) \le p \cdot \mathbf{w}(t) + (\sup p \cdot Y)\beta (t,p)$ and $p \cdot \mathbf{f}'(t) \le p \cdot \mathbf{w}(t) + (\sup p \cdot Y)\beta (t,p)$ a.e. in T, and $\sup p \cdot Y^{S} \le (\sup p \cdot Y)\int _{S}\beta (t,p)dt$ for every $S \in {\mathcal {F}}$. Since $\mathbf{f}(t)$ and $\mathbf{f}'(t)$ are allocations, $\mathbf{f}(t), \mathbf{f}'(t) \in X$ and there exists $y, y' \in Y$ such that $\int _{T}[\mathbf{f}(t) - \mathbf{w}(t)]dt = y$ and $\int _{T}[\mathbf{f}'(t) - \mathbf{w}(t)]dt = y'$. Let $g : {\mathcal {F}} \rightarrow {\mathbb {R}}^{2l+2}_{+}$ be the atomless measure given by

$$\begin{aligned} g(S) = \int _{S} (\mathbf{f}(t), U_{t}(\mathbf{f}(t), p), \mathbf{f}'(t), U_{t}(\mathbf{f}'(t), p))dt \end{aligned}$$

for each coalition $S \in {\mathcal {F}}$. Since 0 and $(x, \alpha , x', \alpha ')$ are values of g, by Liapunov Theorem, there exists $S \in {\mathcal {F}}$ such that

$$\begin{aligned} g(S) = \lambda (x, \alpha , x', \alpha ') + (1 - \lambda )0 = \lambda (x, \alpha , x', \alpha ') = (\lambda x, \lambda \alpha , \lambda x', \lambda \alpha '). \end{aligned}$$

Define

$$\begin{aligned} \mathbf{f}''(t) = \left\{ \begin{array}{ll} \mathbf{f}(t) &{} \quad \text{ for } t \in S,\\ \mathbf{f}'(t), &{} \quad \text{ for } t \not \in S. \end{array}\right. \end{aligned}$$

Then $\mathbf{f}''(t) \in X$ since $\mathbf{f}(t), \mathbf{f}'(t) \in X$, and $\int _{T}{} \mathbf{f}''(t)dt = \int _{S}{} \mathbf{f}(t)dt + \int _{T \setminus S}{} \mathbf{f}'(t) = \lambda x + (1 - \lambda )x' = \lambda \int _{T}\mathbf{f}(t)dt + (1 - \lambda )\int _{T}{} \mathbf{f}'(t)dt$. Since Y is convex,

$$\begin{aligned} \int _{T}[\mathbf{f}''(t) - \mathbf{w}(t)]dt= & {} \int _{T}[\lambda \mathbf{f}(t) + (1 - \lambda )\mathbf{f}'(t) - \mathbf{w}(t)]dt\\= & {} \lambda \int _{T}[\mathbf{f}(t) - \mathbf{w}(t)]dt + (1 - \lambda )\int _{T}[\mathbf{f}'(t) - \mathbf{w}(t)]dt\\= & {} \lambda y + (1 - \lambda )y' \in Y. \end{aligned}$$

Thus, $\mathbf{f}''(t)$ is an allocation. Clearly, we have $p \cdot \mathbf{f}''(t) \le p \cdot \mathbf{w}(t) + (\sup p \cdot Y)\beta (t,p)$ a.e. in T and $\sup p \cdot Y^{S} \le (\sup p \cdot Y)\int _{S}\beta (t,p)dt$ for every $S \in {\mathcal {F}}$. Moreover,

$$\begin{aligned} \int _{T}U_{t}(\mathbf{f}''(t), p)dt = \int _{S}U_{t}(\mathbf{f}(t), p)dt + \int _{T\setminus S}U_{t}(\mathbf{f}'(t), p)dt = \lambda \alpha + (1 - \lambda )\alpha '. \end{aligned}$$

Thus, $\lambda (x, \alpha ) + (1 - \lambda )(x', \alpha ') \in {\mathcal {A}}_{2}(p, y)$, i.e., ${\mathcal {A}}_{2}$ is convex valued. $\Box$