Edgeworth’s conjecture and the number of agents and commodities

Abstract

We answer the question asked by Robert Aumann as to whether core equivalence depends on there being “many more agents than commodities.” We show that for a large class of commodity spaces, which might be infinite-dimensional and even non-separable, core equivalence is indeed equivalent to the presence of “many more agents than commodities” when allocations are Bochner integrable. By contrast, we show that in a classical model of an atomless economy with an infinite-dimensional commodity space, the model where the commodity space is $L_\infty (\mu )$ with the Mackey topology and allocations are Gelfand integrable, core equivalence holds in full generality, even when there are “many more commodities than agents.” The assumptions we make on economies are much weaker than what is commonly used in core equivalence results for infinite-dimensional commodity spaces and reduce to Aumann’s original assumptions when there are finitely many commodities.

Appendices

Appendix 1

(A) Let $E$ be a Banach lattice and let ${\fancyscript{E}}$ be an economy with commodity space $E$. Suppose the consumption set of every agent is $E_{+}$. Now the statement of the notion of “extremely desirable commodity” in Rustichini and Yannelis (1991) says that there are a $v\in E_{+}\!\setminus \! \{0\}$ and a convex solid open neighborhood $U$ of zero in $E$ such that

• (a) for each $t\in T$ and any number $\lambda >0$, $x+\lambda (u+v)\succ _{t} x$ whenever $x\in E_{+}$ and $u\in U$ are such that $x+\lambda (u+v)\in E_{+}$;

• (b) whenever $\delta _{1},\dots ,\delta _{n}$ are positive real numbers and $x_{1},\dots ,x_{n}$ are elements of $E_{+}$ such that $\sum _{i=1}^{n}\delta _{i}=1$ and $x_{i}\notin \delta _{i}U$, $i=1,\dots ,n$, then $\sum _{i=1}^{n}x_{i}\notin U$.¹⁴

Suppose this condition holds. Let $\rho$ be the gauge of $U$. By the properties of $U$, $\rho$ is a Riesz seminorm on $E$, and $U=\{{x\in E}:{\rho (x)<1}\}$. Pick any $x$, $y\in E_{+}$, and set $\alpha =\rho (x)$ and $\beta =\rho (y)$. If $\alpha =0$, then $\rho (x)+\rho (y)=\rho (y)\le \rho (x+y)$, since $\rho$ is a Riesz seminorm. Similarly, if $\beta =0$, then $\rho (x)+\rho (y)\le \rho (x+y)$. Assume that $\alpha$ and $\beta$ are both larger than 0, and set $x_{1}=\frac{1}{\alpha +\beta }x$ and $y_{1}=\frac{1}{\alpha +\beta }y$. Then $x_{1}\notin \frac{\alpha }{\alpha +\beta }U$ and $y_{1}\notin \frac{\beta }{\alpha +\beta }U$. Now by (b), $x_{1}+y_{1}\notin U$, so $\rho (x+y)\ge \alpha +\beta$, and we see again that $\rho (x)+\rho (y)\le \rho (x+y)$. Thus the seminorm $\rho$ is additive on $E_{+}$. There is therefore a positive linear functional $q_{1}$ on $E$ which agrees with $\rho$ on $E_{+}$. By the fact that $U$ is solid, we see that $U=\{{u\in E}:{q_{1}|u|_{}<1}\}$. Moreover, since $E$ is a Banach lattice, the fact that the linear functional $q_{1}$ is positive implies that $q_{1}$ is continuous, i.e., $q_{1}\in E^{*}$. Now it is assumed in Rustichini and Yannelis (1991) that $E$ is separable, so $E^{*}$ has a strictly positive element, $q_{2}$ say (see, e.g., Lindenstrauss and Tzafriri 1979, p. 25). Set $q=q_{1}+q_{2}$. Then $q$ is strictly positive, and $q_{1}|u|_{}<1$ whenever $u\in E$ is such that $q|u|_{}<1$. Hence, by (a), (EDC) must hold.

The other direction, i.e., that (EDC) implies the notion of “extremely desirable commodity” in Rustichini and Yannelis (1991) is immediate.

(B) Let $E$ be a Banach lattice, and let $a$ and $b$ be as in the statement of (US). Choose any $v\in E_{+}$ with $a v>1$. Let $x\in X_{t}$, $u\in E$, and a number $\lambda >0$ be given so that $b|u|_{}<1$ and $x+\lambda (v+u)\in X_{t}$. We may write

$$\begin{aligned} x+\lambda (v+u)=x-\lambda u^{-}+\lambda \bigl (v+u^{+}\bigr ). \end{aligned}$$

Now as $a$ and $b$ are positive, we have

$$\begin{aligned} a(\lambda (v+u^{+}))>\lambda >\lambda b|u|_{}\ge b(\lambda u^{-}). \end{aligned}$$

Thus, (US) implies (EDC).

(C) Let $({\varOmega },{\varSigma },\mu )$ be a $\sigma$-finite measure space, and let the commodity space be $L_{\infty }(\mu )$ with the Mackey topology. Let ${\fancyscript{E}}$ be an economy with probability space of agents $(T,{\fancyscript{T}},\nu )$ (and recall that the measure $\nu$ is complete according to our definition of economy). The assumption in Mertens (1970) on endowments is that the endowment map $e:T\rightarrow L_{\infty }(\mu )_{+}$ is Gelfand integrable (as in (TAE)) and that

• (i) $e(t)\in ||\cdot ||_{\infty }\text {-}{{\mathrm{int}}}L_{\infty }(\mu )_{+}$ for almost all $T$;

• (ii) there is a non-decreasing sequence $\langle {e_{n}}\rangle _{}$ of measurable countably valued functions $e_{n}:T\rightarrow L_{\infty }(\mu )_{+}$ such that $e_{n}(t)\rightarrow e(t)$ in the Mackey topology for almost all $t \in T$.

We will show now that this assumption implies (TAE). Note first that as $(T,{\fancyscript{T}},\nu )$ is a probability space, the property of $e_{n}$ being measurable and countably valued implies that we can find a set $S_{n}\in {\fancyscript{T}}$ with $\nu (T\!\setminus \! S_{n})<2^{-n}$ such that $1_{S_{n}}\times e_{n}$ is a simple function, i.e., takes only finitely many values. Set $T_{n}=\bigcap _{m\ge n}S_{m}$ for each $n$, so that the sequence $\langle {T_{n}}\rangle _{}$ is non-decreasing with $\nu (T_{n})\rightarrow \nu (T)$. Now for each $n$, $1_{T_{n}}\times e_{n}$ is a measurable simple function, and the sequence $\langle 1_{T_{n}}\times e_{n}\rangle _{}$ is non-decreasing such that $(1_{T_{n}}\times e_{n})(t)\rightarrow e(t)$ in the Mackey topology a.e. in $T$.

We may therefore assume that each $e_{n}$ is a simple function. By Fact 3, we may also assume that $({\varOmega },{\varSigma },\mu )$ is a probability space, so that $L_{\infty }(\mu )\subseteq L_{1}(\mu )$. Observe that for almost all $t\in T$, $e_{n}(t)$ is an element of the order interval $[0,e(t)]$ for all $n$. Hence, by Fact 4, we have $e_{n}(t)\rightarrow e(t)$ in $L_{1}(\mu$) for almost all $t\in T$. Thus, as a map from $T$ to $L_{1}(\mu )$, $e$ is strongly measurable, therefore Borel measurable as $(T,{\fancyscript{T}},\nu )$ is complete. Now for each number $r$, the set $\{{x\in L_{1}(\mu )}:{x\ge r1_{{\varOmega }}}\}$ is a closed subset of $L_{1}(\mu )$, and it follows that the set $\{{t\in T}:{e(t)\ge r1_{{\varOmega }}}\}$ is a measurable subset of $T$ for each number $r$.

Note that (i) means that for almost every $t\in T$ there is an integer $n>0$ such that $e(t)\ge (1/n)1_{{\varOmega }}$. By the previous paragraph, we can therefore find a number $r>0$ and a set $H\in {\fancyscript{T}}$ with $\nu (H)\ge r$ such that $e(t)\ge r1_{{\varOmega }}$ for each $t\in H$. For each $n\in \mathbb {N}$, define $e'_{n}:T\rightarrow L_{\infty }(\mu )$ by setting $e'_{n}(t)=e_{n}(t)\vee r1_{{\varOmega }}$ if $t\in H$, and $e'_{n}(t)=e_{n}(t)$ if $t\in T\!\setminus \! H$. Then for each $n$, $e'_{n}$ is a measurable simple function; in particular, $e'_{n}$ is Bochner integrable. Observe that . Thus for each $n$. Moreover, the sequence $\langle {e'_{n}}\rangle _{}$ is non-decreasing, and using the fact that the lattice operations in $L_{\infty }(\mu )$ are continuous for the Mackey topology, we see that $e'_{n}(t)\rightarrow e(t)$ in the Mackey topology for almost all $t\in T$. Thus (TAE) holds.

(D) The following two lemmata show that in the context of Sect. 7, the Gelfand integral does not exhibit pathological features. With $S=T$ and $B={\varOmega }$, Lemma 6 below amounts to a translation of Remark 5 into formal language. Lemma 7 shows that if Assumption (P) holds, and an allocation $f$ is feasible for a coalition $S\in {\fancyscript{T}}$, i.e., , then the agents belonging to $S$ cannot get commodities that are not available in the aggregate endowment of $S$.

Lemma 6

Let $(T,{\fancyscript{T}},\nu )$ be a probability space, $({\varOmega },{\varSigma },\mu )$ a $\sigma$-finite measure space, and $f:T\rightarrow L_{\infty }(\mu )_{+}$ Gelfand integrable. Let $S\in {\fancyscript{T}}$ and write for the Gelfand integral of $f$ over $S$. If a set $B\in {\varSigma }$ is such that $1_{B}\times v=0$, then the set $N=\{t\in S:1_{B}\times f(t)\ne 0\}$ is a null set.

Proof

Choose a strictly positive $q\in L_{1}(\mu )$ (as is possible because $\mu$ is $\sigma$-finite). Note that

As $(1_{B}\times q)(f(t))\ge 0$ for all $t\in S$, it follows that $(1_{B}\times q)(f(t))=0$ for almost all $t\in S$. Consequently, as $(1_{B}\times q)(f(t))=q(1_{B}\times f(t))$ and $q$ is strictly positive, we must have $1_{B}\times f(t)=0$ for almost all $t\in S$. $\square$

Lemma 7

Let $(T,{\fancyscript{T}},\nu )$, $({\varOmega },{\varSigma },\mu )$, $f$, $S$, and, $v$ be as in the previous lemma. Let $C=\{{\omega \in {\varOmega }}:{v(\omega )>0}\}$, identifying $v$ with any of its versions. Then there is a null set $N\subseteq S$ such that $f(t)=1_{C}\times f(t)$ for all $t\in S\!\setminus \! N$.

Proof

Set $B={\varOmega }\!\setminus \! C$, so that $1_{B}\times v=0$. By the previous lemma, there is a null set $N\subseteq S$ such that $1_{B}\times f(t)=0$ for all $t\in S\!\setminus \! N$, which implies that $f(t)=1_{C}\times f(t)$ for all $t\in S\!\setminus \! N$, because $f(t)=1_{B}\times f(t)+1_{C}\times f(t)$. $\square$

Appendix 2

In this appendix, we provide some mathematical background information, collecting some basics on vector integrals and on set theory.

(A) Let $(T,{\fancyscript{T}},\nu )$ be a nontrivial complete and totally finite measure space, and $E$ a Banach space, with norm $\Vert \cdot \Vert$. A measurable function $f{:}\;T\rightarrow E$ is called a simple function if $f(T)$ is finite. The integral of a simple function $f$ is given by

A function $f{:}\;T\rightarrow E$ is strongly measurable if there exists a sequence $\langle f_n\rangle$ of simple functions such that

$$\begin{aligned} \lim _{n\rightarrow \infty }\Vert f(t)-f_n(t)\Vert =0 \end{aligned}$$

for $\nu$-almost all $t$. Every strongly measurable function is Borel measurable. Since for $\nu$-almost all $t$, $f(t)\in {{\mathrm{c\ell }}}\big (\bigcup _n f_n(T)\big )$, a strongly measurable function takes values in a separable subspace of $E$ outside a $\nu$-null set. A strongly measurable function $f{:}\;T\rightarrow E$ is Bochner integrable if there exists a sequence of simple functions $\langle f_n\rangle$ such that

In that case, the Bochner integral of $f$ is

This limit exists and is independent of the particular approximating sequence $\langle f_n\rangle$. A strongly measurable function $f$ is Bochner integrable if and only if the function $t\mapsto \Vert f(t)\Vert$ is Lebesgue integrable (Diestel and Uhl 1977, Theorem II.2.2). Bochner integrals inherit many convenient properties of the Lebesgue integral. If $f$ is Bochner integrable, we call the function from ${\fancyscript{T}}$ to $E$ given by

the indefinite Bochner integral. The indefinite Bochner integral is a $\nu$-continuous countably additive vector measure; $\nu$-continuity meaning that if $\lim _{n\rightarrow \infty }\nu (A_n)=0$, then (Diestel and Uhl 1977, Theorem II.2.4(i)).

In order to be able to integrate functions with values that do not essentially lie in a separable subspace, one can use weak integrals. Let $E^*$ be the topological dual of $E$. A function $f{:}\;T\rightarrow E$ is weakly measurable if $t\mapsto x^*f(t)$ is measurable for all $x^*\in E^*$. A function $f{:}\;T\rightarrow E^*$ is weak*-measurable if $t\mapsto xf(t)$ is measurable for all $x\in E$. By Pettis’ measurability theorem, a function is strongly measurable exactly when it is weakly measurable and there is a separable subspace of $E$ containing almost all values (Diestel and Uhl 1977, Theorem II.1.2). A weak*-measurable function need not be weakly measurable and a weakly measurable function need not be strongly measurable (Diestel and Uhl 1977, Examples II.1.5 and II.1.6). Let $f{:}\;T\rightarrow E$ be weakly measurable. If for each $A\in {\fancyscript{T}}$, there is an $x_{A}\in E$ such that for all $x^*\in E^*$, $f$ is called Pettis integrable and

the Pettis integral of $f$ over $A$. It agrees with the Bochner integral if the latter is well defined, so this notation is unambiguous. It is possible for a weakly measurable function that $t\mapsto x^*f(t)$ is integrable for all $x^*\in E^*$ without $f$ being Pettis integrable (Diestel and Uhl 1977, Example II.3.3). Better behaved in that respect is the Gelfand integral. If $f{:}\;T\rightarrow E^*$ is weak*-measurable and $t\mapsto xf(t)$ is integrable for all $x\in E$, then $f$ is Gelfand integrable and there is a unique $x^*\in E^*$ such that for all $x\in E$ (Diestel and Uhl 1977, Lemma II.3.1). We call

the Gelfand integral of $f$. It follows that whenever $f$ is Gelfand integrable and $A\in {\fancyscript{T}}$, there is an $x_{A}^*\in E^*$ such that for all $x\in E$.

(B) We now collect some set-theoretic results. Many of the results will not be used in proofs, but in discussing how certain results fit into the literature. All results of this paper are derivable from the usual axioms of set theory, i.e., Zermelo–Fraenkel set theory with the axiom of choice (ZFC).

Recall that a partially ordered set is well-ordered if every non-empty subset has a minimum. An ordinal is a set well-ordered by the relation “$\in$ or $=$” and such that every element of the set is also a subset. If $\alpha$ and $\gamma$ are ordinals, we write $\alpha \le \gamma$ if $\alpha \in \gamma$ or $\alpha =\gamma$. Every set of ordinals is well-ordered by $\le$ and each ordinal equals the set of strictly smaller ordinals. In particular, we can use notation such as $\langle x_\xi \rangle _{\xi <\alpha }$ to denote a transfinite sequence indexed by the ordinal $\alpha$. There is no set containing all ordinals.

For each set $X$, there is a smallest ordinal $\kappa$ such that there is a bijection from $\kappa$ onto $X$. We call $\kappa$ the cardinal of $X$ and write $\kappa =\#(X)$. Finite cardinals can be identified with the natural numbers $0,1,2,3,\ldots$ In particular, $2=\{0,1\}$. The set of all finite cardinals is a cardinal itself, the first infinite cardinal $\omega$. The next larger cardinal is $\omega _1$, the first uncountable cardinal. The cardinal of $\mathbb {R}$ is of special importance and is denoted by $\mathfrak {c}$ and called the cardinal of the continuum. If $\kappa$ is a cardinal, we let $\kappa ^+$ be the smallest cardinal strictly larger than $\kappa$. For example, $\omega ^+=\omega _1$. A cardinal of the form $\kappa ^+$ is a successor cardinal and every other cardinal is a limit cardinal. The continuum hypothesis (CH) says that $\omega _1=\mathfrak {c}$.

If $\kappa$ and $\lambda$ are cardinals, we let $\kappa ^\lambda$ be the cardinal of the set of functions from $\lambda$ to $\kappa$. Since one can identify subsets with indicator functions, $2^\kappa$ is the cardinal of the set of all subsets of $\kappa$. For every cardinal $\kappa$, we have $\kappa <2^\kappa$. We have $\mathfrak {c}=2^\omega$. CH can be written as $\omega ^{+}=2^\omega$. The generalized continuum hypothesis (GCH) says that $\kappa ^+=2^\kappa$ for every cardinal $\kappa$. There are other operations one can do with the cardinals $\kappa$ and $\lambda$. We let $\kappa +\lambda$ be the cardinal of the disjoint union of $\kappa$ and $\lambda$ and $\kappa \cdot \lambda$ be the cardinal of their Cartesian product. If $\kappa$ or $\lambda$ is infinite and both are nonzero, these operations are trivial and one has $\kappa +\lambda =\kappa \cdot \lambda =\max \{\kappa ,\lambda \}$.

An axiom A is relatively consistent with ZFC if every proof of a contradiction from ZFC together with A can be turned into a proof of a contradiction from ZFC alone. Clearly, everything is relatively consistent with ZFC if there is a contradiction provable from ZFC. So, we assert our faith in mathematics and assume this is not possible. The generalized continuum hypothesis, and therefore also the continuum hypothesis, is relatively consistent with ZFC (Kunen 2011, Theorem II.6.24). A weakening of the continuum hypothesis is provided by Martin’s axiom. The original statement of Martin’s axiom is slightly intricate, but Martin’s axiom is equivalent to the following statement: If $X$ is a compact Hausdorff topological space in which every disjoint family of non-empty open subsets is countable, then the intersection of less than $\mathfrak {c}$ open dense subsets is non-empty (Kunen 2011, Lemma III.3.17). A consequence of Martin’s axiom is that the additivity of Lebesgue measure is $\mathfrak {c}$, that is, the union of less than $\mathfrak {c}$ Lebesgue null sets is again a null set (Kunen 2011, Lemma III.3.28). Baire’s category theorem for compact Hausdorff spaces shows that Martin’s axiom in its topological version is implied by the continuum hypothesis. However, Martin’s axiom is much weaker; in fact, there is a precise sense in which Martin’s axiom is consistent with the cardinal of the continuum being arbitrarily large (Kunen 2011, Theorem V.4.1).

There are axioms that are widely used and widely taken to be consistent even though their relative consistency cannot be established. For example, one cannot prove the relative consistency of the existence of an atomless probability space in which every subset is measurable. By a result of Ulam, if such a probability space exists, a so-called weakly inaccessible cardinal must exist too (Jech 2003, Theorem 10.1). Such weakly inaccessible cardinals are known to imply the consistency of ZFC (Kunen 2011, Corollary II.6.26 and Theorem II.6.23). But by Gödels second incompleteness theorem, an axiom system strong enough to prove the consistency of ZFC is not relatively consistent with ZFC.