Does the von Neumann-Morgenstern utility theorem work for infinitely many outcomes?

Question

The von Neumann-Morgenstern utility theorem is easy to prove for a finite number of outcomes. Is it still true for an infinite number of outcomes?

With infinite outcomes, a lottery can now be any probability distribution on the set of outcomes.

The problem seems to be that to prove the theorem one would need to use the independence axiom an infinite number of times, which of course is not possible.

If the theorem is false in the infinite case, can we generalize the axioms to make the theorem true?

---

Here is a (sketch of a) proof of the theorem for infinite outcomes if we only consider lotteries with finite support.

Let $A$ be the set of outcomes.

If $X \sim Y$ for every $X, Y \in A$, we define $u$ as $u(Z) = 0$ for all $Z \in A$.

Otherwise, $X \prec Y$ for some $X, Y \in A$. We will define $u(X) = 0$ and $u(Y) = 1$.

For every outcome $Z \in A$, we define $u_Z$ as the von Neumann-Morgenstern utility function for the relation $\prec$ restricted to lotteries over the set $\{X, Y, Z\}$. Since this set is finite, the theorem guarantees $u_Z$ will exist. We also require that $u_Z(X) = 0$ and $u_Z(Y) = 1$, which ensures uniqueness.

We now define $$u(Z) := u_Z(Z)$$ for all $Z \in A$.

Now for any events $L, M$ with finite support, we define the utility function $u_{L,M}$ as the von Neumann-Morgenstern utility function for $\prec$ restricted to the set $\{X, Y\} \cup \operatorname{supp}(L) \cup \operatorname{supp}(M)$. We again require that $u_{L,M}(X) = 0$ and $u_{L,M}(Y) = 1$. Now for every $Z$ in $u_{L,M}$'s domain, $u_Z$ is just a restriction of $u_{L,M}$. Therefore $u_{L,M}(Z) = u_Z(Z) = u(Z)$.

$L \preceq M$ iff $E[u_{L,M}(L)] \le E[u_{L,M}(M)]$ iff $E[u(L)] \le E[u(M)]$

Q.E.D.

Answer 1

It is not fully clear to me what the question is, but there is a huge literature on versions of the von Neumann and Morgenstern utility representation on infinite sets. The most common approach is to impose a continuity axiom. (Formally, there are different alternatives, f.i.\ assuming an archimedean property.)

Among many possible references, I recommend J.-M. Grandmont, "Continuity properties of a von Neumann-Morgenstern utility", Journal of Economic Theory 4, 45-57. Theorem 2 provide simple conditions for existence that elucidate the role of the independence (or linearity) axiom. And Theorem clarifies the difference between assuming a continuous representation or a continuous NM-utility function.

Answer 2

No, it does not hold in the infinite case (as stated). To see this, we will define a VNM-rational $\prec$ that does not have a utility function.

Let $A$ be any infinite set of outcomes. Then for lotteries $L$ and $M$, $L \prec M$ iff the support of $L$ is finite and of $M$ is infinite.

Completeness: We define $L \preceq M$ as $M \nprec L$. It is clear that $M \nprec L$ or $L \nprec M$, since otherwise $L \prec M$ and $M \prec L$, implying $L$ has both finite and infinite support, a contradiction.

Transitivity: If $L \prec M$ and $M \prec N$, $L$ has finite support and $N$ has infinite support. Ergo, $L \prec N$.

Continuity: Given $L \preceq M \preceq N$, then we have two cases:

• The support of $M$ is finite. Then $M \sim 1L + 0N$.
• The support of $M$ is infinite. Then $M \sim 0L + 1N$.

Independence: If $L \preceq M$ and $p \in [0,1]$, there are four cases:

• $p = 0$: $0L + 1N \sim 0M + 1 N$
• $0 < p < 1$ and the support of $N$ is finite: $pL + (1-p)N \sim L \preceq M \sim pM + (1-p)N$
• $0 < p < 1$ and the support of $N$ is infinite: $pL + (1-p)N \sim pM + (1-p)N$
• $p = 1$: $1L + 0N \preceq 1M + 0N$

So $\prec$ satisfies the VNM-axioms. However, there is no utility function $u$ such that $L \prec M \iff u(L) < u(M)$. To see this, note that $u$ must assign every outcome to some value $f$, since if $L=1x$ and $M=1y$ then

$$L \sim M$$ $$E[u(L)] = E[u(M)]$$ $$u(x) = u(y)$$

For any lottery $N$ with infinite support (which exists since $A$ is infinite), $E[u(N)]$ also equals $f$, since $u$ assigns each outcome in $L$'s support to $f$. So $E[u(L)] = E[u(N)]$. This implies $L \sim N$, which is false since $L$'s support is finite and $N$'s support is infinite.

$\Box$