WHY SATURATED PROBABILITY SPACES ARE NECESSARY

H. JEROME KEISLER AND YENENG SUN

Abstract. An atomless probability space (Ω, A, P) is said to have the saturation property for a probability measure μ on a product of Polish spaces X × Y if for every random element f of X whose law is marg_{X}(μ), there is a random element g of Y such that the law of (f, g) is μ. (Ω, A, P) is said to be saturated if it has the saturation property for every such μ. We show each of a number of desirable properties holds for every saturated probability space and fails for every non-saturated probability space. These include distributional properties of correspondences, such as convexity, closedness, compactness and preservation of upper semi-continuity, and the existence of pure strategy equilibria in games with many players. We also show that any probability space which has the saturation property for just one “good enough” measure, or which satisfies just one “good enough” instance of the desirable properties, must already be saturated.

Our underlying themes are: (1) There are many desirable properties that hold for all saturated probability spaces but fail everywhere else; (2) Any probability space that out-performs the Lebesgue unit interval in almost any way at all is already saturated.

1. Introduction

Atomless probability spaces are widely used in mathematics and its applications. However, it has been found in [8], [14], [15] and [24] that the typical atomless probability space, the Lebesgue unit interval, does not have a number of desirable properties. Properties that fail on the Lebesgue unit interval include saturation properties for random elements and stochastic processes, existence of solutions of stochastic integral equations, regularity properties for distributions of correspondences such as convexity, closedness, compactness and preservation of upper semi-continuity, and the existence of pure strategy equilibria in games with many players.

In the papers [8], [15] and [24] it was shown that there do exist atomless probability spaces with these desirable regularity properties. These earlier papers used methods from nonstandard analysis. In this paper, using ordinary standard methods, we find exactly which probability spaces have these properties—each of these properties holds for every saturated probability space, and fails for every nonsaturated atomless probability space. This improves the earlier results.

Formally, an atomless probability space (Ω, A, P) is said to have the saturation property for a probability measure μ on a product of Polish spaces X × Y if for every random element f of X whose law is marg_{X}(μ), there is a random element g of Y such that the law of (f, g) is μ. (Ω, A, P) is said to be saturated if it has the saturation property for every such μ.

Date: February 18, 2009.

We gratefully acknowledge support from the Vilas Trust Fund and the NUS Research Grant R-146-000-082-112.

Saturated probability spaces were introduced in [8]. It was shown there that a probability space (Ω, A, P) is saturated if and only if it satisfies the following condition: The measure P restricted to a set of positive measure is never countably generated modulo the null sets.^{1} By Maharam’s theorem, a probability space satisfies this condition if and only if its measure algebra is a finite or countable convex combination of measure algebras of uncountable powers of [0, 1].

We will first prove local saturation results showing that if (Ω, A, P) has the saturation property for just one “good enough” measure on a product of Polish spaces then Ω must already be saturated. We then will apply the saturation property directly to give characterizations of saturated probability spaces by distributional properties of correspondences, and by the existence of pure strategy equilibria in games with many players. We will then apply the local saturation results to show that one “good enough” instance of one of these properties already implies that the space is saturated.

Our underlying themes are: (1) There are many desirable properties that hold for all saturated probability spaces but fail everywhere else; (2) Any probability space that out-performs the Lebesgue unit interval in almost any way at all is already saturated.

The paper is organized as follows. Fundamental results about saturated probability spaces are presented in Section 2, including a brief review of global conditions for saturation and new local conditions for saturation. Section 3 deals with various distributional properties of correspondences by saturation, while Section 4 concerns the existence of pure strategy equilibria in games with many players. In Section 5, we point out that if these results on the distributional properties of correspondences and the existence of pure strategy equilibria in large games are established for one particular saturated probability space, then it follows easily that they hold for any other saturated probability space.

While the notions and methods here are entirely standard, this paper owes a great deal to earlier work using nonstandard methods, in particular using the Loeb probability spaces introduced in [17]. The papers [15] and [24] showed that atomless Loeb spaces have the desired properties for correspondences and large games. One realized that the atomless Loeb spaces are very rich in the sense that they have many more measurable sets than the Lebesgue unit interval. The richness is fully captured by the standard notion of a saturated probability space—it was shown in [8] that every atomless Loeb space is saturated. This gave a hint that these properties might hold for all saturated probability spaces. The earlier work with Loeb spaces was the motivation for Section 5 in this paper. The known facts that the desired properties hold for atomless Loeb spaces and that atomless Loeb spaces are saturated, combined with Section 5 here, give an alternative proof that the desired properties hold for all saturated probability spaces.

The positive direction of our results here can be compared with results in the literature on stochastic integral equations. In the papers [9], [10], and [16], it was shown that adapted Loeb probability spaces have the desirable property that every stochastic integral equation with random continuous coefficients has a strong solution. In the paper [8] it was shown by standard methods that every saturated adapted probability space has this desirable property. In the papers [3] and [4], a variety of other existence and optimization results for stochastic integrals and controls were obtained for all saturated adapted probability spaces (using [11]).

1This condition is called “ℵ_{1}-atomless” in [8], “nowhere countably generated” in [18], “rich”

in [21] and our earlier draft [13], and “super-atomless” in [22].

In this paper we work with ordinary probability spaces, while the earlier work on stochastic integral equations involved adapted probability spaces. The notion of a saturated adapted probability space was introduced in [8], where it was shown that every adapted Loeb probability space is saturated, and that every saturated adapted probability space is saturated as a probability space. An important difference between this paper and the earlier work on stochastic integral equations is that here we also get converse results showing that the desired properties fail on every non-saturated probability space, and we get even stronger local converse results.

2. Saturated Probability Spaces

In what follows, X, Y, Z, … denote Polish spaces (complete metrizable topological spaces), and M(X) is the space of all Borel probability measures on X with the Prohorov metric ρ. We recall that M(X) is again a Polish space, M(X) has the topology of weak convergence, and if X is compact then so is M(X). For each μ ∈ M(X × Y), let marg_{X}(μ) be the marginal of μ in M(X); thus marg_{X} : M(X × Y) → M(X) is a continuous surjection. Throughout this paper, probability space means complete countably additive probability space. The triples (Ω, A, P), (Γ, C, Q) will denote atomless probability spaces. (T, L, λ) is the usual Lebesgue probability space on the unit interval T = [0, 1]. L^{0}(Ω, X) is the space of all random elements of X (measurable functions f : Ω → X) with the metric of convergence in probability. The law (or distribution) function law : L^{0}(Ω, X) → M(X) is defined by law (f)(U) = P (f^{−1}(U)) for each Borel set U.

If A is a σ-algebra and S ∈ A, we let A^{S} = {B ∈ A : B ⊆ S}. The set of all P -null sets is denoted by N, or N (P). We use f, g to denote random elements of X, Y respectively, on some probability space (Ω, A, P). Given a random element f, σ(f) is the smallest σ-algebra that contains {f^{−1}(U) : U Borel} ∪ N.

We will often use the following well-known facts about the law mappings on an arbitrary probability space (Ω, A, P).

Lemma 2.1. (i) law : L^{0}(Ω, X) → M(X) is continuous.

(ii) If (Ω, A, P) is atomless, then law : L^{0}(Ω, X) → M(X) is surjective.

(iii) Suppose law (f_{n}) weakly converges to ν and law (g_{n}) weakly converges to π as n → ∞. Then some subsequence of law (f_{n}, g_{n}) weakly converges to a measure μ ∈ M(X × Y) such that marg_{X}(μ) = ν and marg_{Y} (μ) = π.

Proof. We prove (iii). The sequence law (f_{n}) is relatively compact, since it is contained in the compact set {law (f_{n}) : N ∈ ℕ} ∪ {ν}. By the converse Prohorov’s theorem (see [1]), law (f_{n}) is tight, that is, for each ε > 0 there is a compact set J_{ε} in X such that P (f_{n}(ω) ∈ J_{ε}) ≥ 1 − ε for each n. Similarly, there is a compact set K_{ε} in Y such that P (g_{n}(ω) ∈ K_{ε}) ≥ 1 − ε for each n. Then J_{ε} × K_{ε} is compact in X × Y, and P ((f_{n}, g_{n})(ω) ∈ J_{ε} × K_{ε}) ≥ 1 − 2ε for each n. Thus the sequence law (f_{n}, g_{n}) is tight. By the direct Prohorov theorem, the sequence {law (f_{n}, g_{n}) : n ∈ ℕ} is contained in a compact set C ⊆ M(X × Y). Therefore some subsequence of law (f_{n}, g_{n}) weakly converges to a measure μ ∈ M(X × Y). Since the functions marg_{X} and marg_{Y} are continuous, it follows that marg_{X}(μ) = ν and marg_{Y} (μ) = π.

Definition 2.2. (i) (Ω, A, P) is said to satisfy the saturation property for a measure μ ∈ M(X × Y) if for every f ∈ L^{0}(Ω, X) with law (f) = marg_{X}(μ), there exists g ∈ L^{0}(Ω, Y) such that law (f, g) = μ.

(ii) A probability space (Ω, A, P) is saturated (or has full saturation) if (Ω, A, P) is atomless, and for every X, Y, (Ω, A, P) satisfies the saturation property for every μ ∈ M(X × Y).

Note that for a pair of random elements (f′, g′) on some probability space, the space (Ω, A, P) satisfies the saturation property for law (f′, g′) if and only if for every f ∈ L^{0}(Ω, X) with law (f) = law (f′), there exists g ∈ L^{0}(Ω, Y) with law (f, g) = law (f′, g′).

As shown in [8], typical examples of saturated probability spaces are atomless Loeb probability spaces, and product spaces of the form {0, 1}^{κ} and [0, 1]^{κ}, where κ is an uncountable cardinal, {0, 1} has the uniform measure, and [0, 1] has the Lebesgue measure. The Lebesgue unit interval (T, L, λ) and the probability spaces [0, 1]^{ℕ} and {0, 1}^{ℕ} are examples of atomless probability spaces that are not saturated.

The following two propositions deal with the trivial case that every atomless probability space has the saturation property for law (f, g).

Proposition 2.3. If f has countable range, then every atomless probability space has the saturation property for law (f, g).

Proof. Let law (f′) = law (f). By modifying f′ on a null set, we may assume that f′ has the same range as f. If f is a constant function, so is f′, and Lemma 2.1 gives us a g′ such that law (f′, g′) = law (f, g). In the general case, f′ is the union of countably many constant functions, and we obtain a g′ with law (f′, g′) = law (f, g) by taking a countable union of functions that work on each constant part of f′.

In view of Proposition 2.3, in the rest of this section we will concentrate on the case that law (f) is an atomless measure.

Proposition 2.4. Suppose law (f) is atomless. The following are equivalent: (i) The Lebesgue unit interval (T, L, λ) has the saturation property for law (f, g). (ii) Every atomless probability space has the saturation property for law (f, g). (iii) g is σ(f)-measurable.

Proof. It is clear that (iii) implies (ii) and that (ii) implies (i). We assume (i) and prove (iii). Since law (f) is atomless and [0, 1] is separable, there is an f′ ∈ L^{0}([0, 1], X) such that law (f′) = law (f) and σ(f′) is the set of all Borel subsets of [0, 1]. By saturation there exists g′ ∈ L^{0}([0, 1], Y) such that law (f′, g′) = law (f, g). Then g′ must be σ(f′)-measurable, so g is σ(f)-measurable.

2.1. Global Conditions for Saturation. In this subsection we give some global necessary and sufficient conditions for a probability space to be saturated. We first list some results from [8] and [19].

Fact 2.5. For each atomless probability space (Ω, A, P), the following are equivalent:

(i) (Ω, A, P) is saturated.

(ii) There is no set S ∈ A such that P (S) > 0 and A^{S} is countably generated modulo the null sets.

(iii) The measure algebra of (Ω, A, P) is a finite or countable convex combination of measure algebras of the form [0, 1]^{κ} where κ is an uncountable cardinal.

(iv) If f ∈ L^{0}(Ω, X), g_{n} ∈ L^{0}(Ω, Y) for each n ∈ ℕ, and law (f, g_{n}) converges weakly to a measure μ, there exists g ∈ L^{0}(Ω, Y) such that law (f, g) = μ.

Proof. The equivalence of (i) and (ii) is proved in [8], Corollary 4.5. The equivalence of (ii) and (iii) follows from Maharam’s Theorem ([19]). A direct proof that (i) is equivalent to (iii) is also given in [5], Theorem 3B.7. The equivalence of (i) and (iv) is a special case of Theorem 5.2 in [8], and is proved directly in [12], Proposition 2.3.

We will often use condition (iv). This condition says that if a law problem has approximate solutions, then it has an exact solution. Some additional global conditions for full saturation are given in [11] (Theorem 4.8), and in [22]. The papers [3], [4], [8], [11], and the monographs [5] and [10], studied the more complicated saturated adapted probability spaces, as well as saturated probability spaces.

Here is a global characterization of full saturation that appears to be new.

Proposition 2.6. An atomless probability space (Ω, A, P) is saturated if and only if:

(v) For each X, Y, compact set C ⊆ M(X × Y), and f ∈ L^{0}(Ω, X), the set

{law (g) : g ∈ L^{0}(Ω, Y) and law (f, g) ∈ C}

is compact.

Proof. It is shown in [12], Proposition 4.9, that condition (iv) of Fact 2.5 implies (v).

We prove that (v) implies condition (iv) of Fact 2.5. If Y has only one point, then so does L^{0}(Ω, Y), and (iv) is trivially true. Assume that Y has at least two points, and (v) holds. Suppose that law (f, g_{n}) converges weakly to μ in M(X ×Y). Let ν = marg_{Y} (μ) ∈ M(Y). Since (Ω, A, P) is atomless and Y has more than one point, we can perturb g_{n} to a sequence h_{n} in L^{0}(Ω, Y) such that law (f, h_{n}) converges weakly to μ and in addition that law (h_{n}) ≠ ν for each n. Let

C = {law (f, h_{n}) : n ∈ ℕ} ∪ {μ}.

Then C is compact in M(X × Y). By (v) the set

D = {law (g) : g ∈ L^{0}(Ω, Y) and law (f, g) ∈ C}

is compact in M(Y). For each n, law (f, h_{n}) ∈ C and hence law (h_{n}) ∈ D. Moreover, law (h_{n}) = marg_{Y} (law (f, h_{n})), and since marg_{Y} is continuous, law (h_{n}) converges weakly to ν. Therefore ν ∈ D, and thus there exists g ∈ L^{0}(Ω, Y) such that law (f, g) ∈ C and law (g) = ν. Since law (h_{n}) ≠ law (g), we have law (f, g) ≠ law (f, h_{n}) for each n. Hence law (f, g) = μ. This proves (iv).

2.2. Local Conditions for Saturation. We will now begin to address our underlying theme (2), that any probability space that out-performs the Lebesgue unit interval in almost any way is saturated.

The following theorem shows that one particular non-trivial instance of the saturation property already implies full saturation. This gives a condition for full saturation that is local in the sense that it involves one particular measure on the product X × Y.

Theorem 2.7. Let f, g be random elements of X, Y on some probability space and assume that law (f) is atomless. Suppose (Ω, A, P) has the saturation property for law (f, g) but the Lebesgue unit interval (T, L, λ) does not. Then (Ω, A, P) is saturated.

Proof. We assume that (Ω, A, P) is not saturated and arrive at a contradiction.

We may take f, g to be random elements on (Ω, A, P). Since (T, L, λ) does not have the saturation property for law (f, g), g is not σ(f)-measurable. By Fact 2.5 (ii), there is an n ∈ ℕ and a set S ∈ A of measure P (S) = 1/n such that A^{S} is countably generated (modulo the null sets). Using the fact that (Ω, A, P) is atomless, there is a countably generated σ-algebra A_{0} such that A^{S} ⊆ A_{0} ⊆ A and A_{0} is atomless with respect to P. Since law (f) is atomless, we may partition Ω into n σ(f)-measurable subsets U_{1}, …, U_{n} of measure 1/n. Because g is not σ(f)-measurable, there is at least one of the sets U = U_{k} such that the restriction of g to U is not σ(f)-measurable.

By Maharam’s theorem [19], any two countably generated atomless measure algebras are isomorphic. Since P (S) = P (U), it follows that there is a measure algebra isomorphism ψ from the measure algebra of (Ω, σ(f), P) to the measure algebra of (Ω, A_{0}, P) such that ψ(U/N) = S/N. By [6], Theorem 4.12 on page 937, ψ is induced by a measurable mapping h : Ω → Ω. Then f′(ω) = f(h(ω)) is measurable. We have σ(f′) = A_{0} and law (f′, 1_{S}) = law (f, 1_{U}), where 1_{S} and 1_{U} are the respective indicator functions of S and U. By the saturation property for law (f, g), there exists g′ on (Ω, A, P) such that law (f′, g′) = law (f, g). Since U is σ(f)-measurable, and law (f′, 1_{S}) = law (f, 1_{U}), there is a Borel measurable function φ such that 1_{U} = φ(f) and 1_{S} = φ(f′). Hence, we have law (f′, g′, 1_{S}) = law (f, g, 1_{U}). The restriction of g to U is not σ(f)-measurable, but the restriction of g′ to S is σ(f′)-measurable because A^{S} ⊆ σ(f′). This contradicts the fact that law (f, g) = law (f′, g′). Therefore (Ω, A, P) is saturated after all.

As a corollary, we see that only trivial cases of the saturation property hold for a probability space that is not saturated. This corollary is a generalization of Proposition 2.4

Corollary 2.8. Suppose (Ω, A, P) is atomless but not saturated, and let f, g be random elements of X, Y on some probability space such that law (f) is atomless. Then (Ω, A, P) has the saturation property for law (f, g) if and only if g is σ(f)- measurable.

Proof. By Proposition 2.4 and Theorem 2.7.

When the space (Ω, A, P) is clear from the context, law^{−1}(ν) will denote the set of all f ∈ L^{0}(Ω, X) such that law (f) = ν. Given a set C ⊆ M(X × Y) and a random element f ∈ L^{0}(Ω, X), let

C(f) = C ∩ {law (f, g) : g ∈ L^{0}(Ω, Y)}.

By Fact 2.5 (iv), on a saturated probability space the set C(f) is closed for every closed set C and random element f of X.

Theorem 2.9. Assume that C ⊆ M(X × Y) and ν = marg_{X} μ for some μ ∈ C. Suppose that C(f) is closed for every f ∈ law^{−1}(ν) on (Ω, A, P), but there is an f′ ∈ law^{−1}(ν) on the Lebesgue unit interval (T, L, λ) such that C(f′) is not closed and σ(f′) = L. Then (Ω, A, P) is saturated.

Proof. Since σ(f′) = L, ν is an atomless measure. Let μ ∈ cl(C(f′)) \ C(f′). Then there is a sequence g′_{n} ∈ L^{0}(T, Y) such that law (f′, g′_{n}) = μ_{n} ∈ C and μ_{n} converges weakly to μ. Since σ(f′) = L, each g′_{n} is σ(f′) measurable, and hence there is a Borel function ψ_{n} : X → Y such that g′_{n} = ψ_{n}(f′) a.s. Let f ∈ law^{−1}(ν) on (Ω, A, P), and let g_{n} = ψ_{n}(f). Then law (f, g_{n}) = law (f′, g′_{n}) = μ_{n}, so μ_{n} ∈ C(f). Since C(f) is closed, μ ∈ C(f). Hence there exists g ∈ L^{0}(Ω, Y) with law (f, g) = μ. It follows that (Ω, A, P) has the saturation property for μ but (T, L, λ) does not. By Theorem 2.7, (Ω, A, P) is saturated.

The following lemma is a sort of approximate saturation that holds for every atomless probability space.

Lemma 2.10. Suppose μ ∈ M(X × Y), ν = marg_{X} μ, and ν is atomless. Then for each f ∈ law^{−1}(ν) there is a sequence of σ(f)-measurable random elements g_{n} of Y such that law (f, g_{n}) converges weakly to μ.

Proof. Take (f′, g′) ∈ law^{−1}(μ). There is a sequence of σ(f′)-measurable simple functions f′_{n} in L^{0}(Ω, X) such that f′_{n} converges to f′ a.s. Since law (f′) is atomless, for each n there is a σ(f′)-measurable random element g′_{n} of Y such that law (f′_{n}, g′_{n}) = law (f′_{n}, g′). Since f′_{n} → f′ a.s., law (f′_{n}, g′) converges weakly to law (f′, g′) = μ. Because each g′_{n} is σ(f′)-measurable, there are σ(f)-measurable g_{n} such that law (f, g_{n}) = law (f′, g′_{n}), and law (f, g_{n}) converges weakly to μ.

The next theorem is a consequence of Theorem 2.9 and Lemma 2.10.

Theorem 2.11. Suppose ν is an atomless measure in M(X), Y has cardinality greater than 1, and C = M(X × Y). Then (Ω, A, P) is saturated if and only if for each f ∈ law^{−1}(ν), the set C(f) is closed in M(X × Y).

Proof. We prove the non-trivial direction. Suppose C(f) is closed for every f ∈ law^{−1}(ν). Take two distinct points y_{0}, y_{1} in Y. There is a random element (f, g) of X × Y such that f ∈ law^{−1}(ν), g ∈ L^{0}(Ω, {y_{0}, y_{1}}), P (g^{−1}({y_{1}}) = 1/2, and g is independent of σ(f). Then g is not σ(f)-measurable. Let μ = law (f, g). Since ν is atomless there is an f′ ∈ L^{0}(T, X) such that law (f′) = ν and σ(f′) = L. Then μ ∉ C(f′). By Lemma 2.10, μ belongs to the closure of C(f′). Hence C(f′) is not closed. By Theorem 2.9, (Ω, A, P) is saturated.

3. Distribution of Correspondences on Saturated Probability Spaces

Measurable correspondences and their selections are important in many areas of mathematics, including optimization, control theory, pattern analysis, stochastic analysis, and mathematical economics. The paper [24] developed a theory of distribution of correspondences on probability spaces, and proved that Loeb probability spaces have several desirable regularity properties that fail for the more familiar probability spaces such as the Lebesgue unit interval. We’ll call these properties P 1–P 6. The proofs of P 1–P 6 for Loeb spaces in [24] made substantial use of methods from nonstandard analysis. In this section we will prove, without using nonstandard methods, that each saturated probability space has each property P 1–P 6, and for each non-saturated probability space, each property P 1–P 6 fails.

In Subsection 3.2 we apply the full saturation condition directly to prove that every saturated probability space has properties P 1–P 6. In Subsection 3.3 we prove global converse results, showing that each of the properties P 1–P 6 fails for every non-saturated probability space. These two subsections address our underlying theme (1), that many desirable properties hold for all saturated probability spaces but fail everywhere else. In Subsection 3.4 we use Theorem 2.7 to prove stronger local converse results, which show that a single “good enough” instance of one of the properties P 1–P 4 or P 6 already implies full saturation. This addresses our underlying theme (2), that any probability space that out-performs the Lebesgue unit interval in almost any way at all is already saturated.

3.1. Background. We refer to [24], Section 2, for some basic standard notions and results on correspondences. Here is a brief summary.

By definition, every Polish space is separable and admits a complete metric. Since a Polish space is compact if and only if it admits a complete totally bounded metric, and every Polish space is embeddable in the compact Polish space [0, 1]^{ℕ}, every Polish space admits a (not necessarily complete) totally bounded metric (see pages 217–219 of [1]).

Throughout this section we let X be a Polish space and let d be a totally bounded metric on X.

A correspondence from Y to Z is a mapping from Y to the family of nonempty subsets of Z. Let G be a correspondence from a probability space (Ω, A, P) to X. A measurable mapping g : Ω → X is called a measurable selection of G if g(ω) ∈ G(ω) for P -almost all ω ∈ Ω. The correspondence G is said to be measurable if its graph

{(ω, x) ∈ Ω × X : x ∈ G(ω)}

belongs to the product σ-algebra A⊗B(X), where B(X) denotes the Borel σ-algebra on X.

The correspondence G is said to be closed (compact) valued if G(ω) is a closed (compact) subset of X for all ω ∈ Ω. For each set B ⊆ X, define

G^{−1}(B) = {ω : G(ω) ∩ B ≠ ∅}.

If G is closed valued, then G is a measurable correspondence if and only if G^{−1}(O) is measurable for every open set O in X.

For a point x ∈ X and a nonempty subset B of X, let the distance d(x, B) from the point x to the set B be inf{d(x, y) : y ∈ B}. For nonempty subsets A and B of X, the corresponding Hausdorff distance ρ_{d}(A, B) between the sets A and B is defined by

ρ_{d}(A, B) = max{sup_{x∈A} d(x, B), sup_{y∈B} d(y, A)}.

Let F_{X} be the hyperspace of nonempty closed subsets of X, equipped with the metric ρ_{d}.

We will need the following elementary facts from [24], which allow us to treat measurable correspondences as mappings into a new Polish space and to apply full saturation to such mappings.

Fact 3.1. (i) The hyperspace F_{X} with the Hausdorff distance ρ_{d} is still a Polish space.

(ii) Let G be a closed valued measurable correspondence from a probability space (Ω, A, P) to X. Then the induced mapping ω ↦ G(ω) is a measurable function from (Ω, A, P) to M(F_{X}).

Proof. See [24], Lemma 2.2 and Proposition 2.3.

Definition 3.2. Let G be a correspondence from (Ω, A, P) to X.

(i) The distribution of G is the set

D_{G} = {law (g) : g is a measurable selection of G}.

(ii) Suppose G is a closed valued measurable correspondence. The law of G is the induced measure law (G) ∈ M(F_{X}) defined by law (G)(U) = P (G^{−1}(U)).

In the above definition, the topology on F_{X} and the law of a correspondence depend on the metric d on X, not just on the topology of X.

Definition 3.3. A correspondence G from a topological space Y to another topological space Z is said to be upper semi-continuous at y_{0} ∈ Y if for any open set U that contains G(y_{0}), there exists a neighborhood V of y_{0} such that y ∈ V implies that G(y) ⊆ U. G is upper semi-continuous if it is upper semi-continuous at every point y ∈ Y.

For a compact valued correspondence, upper semi-continuity has the following characterization by sequences (see [7], page 24).

Fact 3.4. Let G be a compact valued correspondence from a Polish space Y to a Polish space Z. G is upper semi-continuous at a point y ∈ Y if and only if whenever y_{n} converges to y in Y and z_{n} ∈ G(y_{n}) for each n, z_{n} has a subsequence that converges to a point z ∈ G(y).

We need the following approximate version of Proposition 3.5 of [24]. This approximate version holds for all atomless probability spaces and has an elementary proof.

Proposition 3.5. Let F be a closed valued measurable correspondence from an atomless probability space (Ω, A, P) to a Polish space X, and let μ be a Borel probability measure on X. The following are equivalent.

(i) μ belongs to the closure of D_{F}.

(ii) For every open set O in X, μ(O) ≤ P (F^{−1}(O)).

Proof. (i) ⇒ (ii): Assume (i). There is a sequence f_{n} of measurable selections of F such that μ_{n} = law (f_{n}) weakly converges to μ. For each n ∈ ℕ and open set O in X, f_{n}^{−1}(O) ⊆ F^{−1}(O), so μ_{n}(O) ≤ P (F^{−1}(O)). Hence μ(O) ≤ lim inf_{n→∞} μ_{n}(O) ≤ P (F^{−1}(O)).

(ii) ⇒ (i): The proof is the same as the proof of the implication (iv) ⇒ (i) in Proposition 3.5 of [24], but it uses an arbitrary atomless probability space instead of a Loeb space.

3.2. Regularity Properties for Distribution of Correspondences. In this subsection we generalize the main theorems of [24] to all saturated probability spaces. We let X, Y be Polish spaces.

Given a closed valued measurable correspondence F from (Ω, A, P) to X, we say that D_{F} is maximal if we have D_{F} ⊇ D_{G} for every closed valued measurable correspondence G that has the same law as F.

Theorem 3.6. Let (Ω, A, P) be a saturated probability space. Then we have the following.

P 1: For each closed valued measurable correspondence F from (Ω, A, P) to X, D_{F} is maximal.

P 2: For any correspondence F from (Ω, A, P) to X, D_{F} is convex.

P 3: For any closed valued correspondence F from (Ω, A, P) to X, D_{F} is closed.

P 4: For any compact valued correspondence F from (Ω, A, P) to X, D_{F} is compact.

P 5: Let F be a compact valued correspondence from (Ω, A, P) to X. Suppose that Y is a metric space and G is a closed valued correspondence from Ω × Y to X such that:

(a) For all (ω, y) ∈ Ω × Y, G(ω, y) ⊆ F (ω).

(b) For each fixed y ∈ Y, G(·, y) (denoted by G_{y}) is a measurable correspondence from (Ω, A, P) to X.

(c) For each fixed ω ∈ Ω, G(ω, ·) is upper semi-continuous from Y to X.

Then the correspondence H(y) = D_{G_{y}} is upper semi-continuous from Y to M(X).

P 6: Let G be a measurable mapping from (Ω, A, P) to the space M(X) of probability measures on X. Then there is a measurable mapping f from (Ω, A, P) to X such that:

(a) for every Borel set B in X, law (f)(B) = ∫_{Ω} G(ω)(B) dP;

(b) for each ω ∈ Ω, f(ω) ∈ supp G(ω), where supp G(ω) is the support of the probability measure G(ω) on X.

Proof. P 1: Suppose law (G) = law (F) and g is a measurable selection of G. Let

H = {(x, A) ∈ X × F_{X} : x ∈ A}.

Since g is a measurable selection of G, we have law (g, G)(H) = 1. By full saturation there is an f ∈ L^{0}(Ω, X) such that law (f, F) = law (g, G). Then law (f, F)(H) = 1, which means that f is a measurable selection of F. Therefore, μ = law (f) = law (g) ∈ D_{F}. This shows that D_{G} ⊆ D_{F}.

P 2: Let μ, ν ∈ D_{F}. Then there are measurable selections f, g of F such that law (f) = μ and law (g) = ν. Let G(ω) = {f(ω), g(ω)}. Then G ⊆ F and G is a closed valued measurable correspondence from (Ω, A, P) to X. Let α ∈ (0, 1). We show that there is a measurable selection h of G such that

law (h) = αμ + (1 − α)ν.

Choose sequences of simple functions f_{n}, g_{n} : Ω → X such that f_{n}(ω) is within 2^{−n} of f(ω), and g_{n}(ω) is within 2^{−n} of g(ω), with probability at least 1 − 2^{−n}. For each n there is a finite measurable partition P_{n} of Ω such that:

(a) For each S ∈ P_{n}, f_{n} and g_{n} are constant on S;

(b) The union of the sets S ∈ P_{n} on which f_{n} is everywhere within 2^{−n} of f and g_{n} is everywhere within 2^{−n} of g has probability at least 1 − 2 · 2^{−n}.

Since (Ω, A, P) is atomless, for each S ∈ P_{n} there is a measurable set S_{0} ⊆ S such that P (S_{0}) = αP (S). Let h_{n} : Ω → X be the simple function such that on each partition set S ∈ P_{n}, h_{n}(ω) = f_{n}(ω) for any ω ∈ S_{0}, and h_{n}(ω) = g_{n}(ω) for any ω ∈ S \ S_{0}. Then

law (h_{n}) = αlaw (f_{n}) + (1 − α)law (g_{n}).

Moreover, h_{n}(ω) is within 2^{−n} of G(ω) with probability at least 1 − 2 · 2^{−n}, and law (h_{n}) converges weakly to αμ + (1 − α)ν. Since the law function is continuous, law (f_{n}, g_{n}) also converges weakly to law (f, g). By Lemma 2.1, some subsequence of law (f_{n}, g_{n}, h_{n}, G) converges weakly to a measure π in M(X × X × X × F_{X}) whose marginals on the three copies of X are respectively μ, ν, and αμ+(1−α)ν. Without loss of generality, we assume that the sequence law (f_{n}, g_{n}, h_{n}, G) converges weakly to π. Let τ be the marginal measure of π on the third copy of X with F_{X}. Then, law (h_{n}, G) converges weakly to τ. For each k ≥ 1 let H^{k} = {(x, A) ∈ X × F_{X} : d(x, A) ≤ 1/k}. Then

τ(H^{k}) ≥ lim sup_{n→∞} law (h_{n}, G)(H^{k}) = 1,

which implies that τ(H^{k}) = 1. Therefore, τ(H) = 1. Since (Ω, A, P) is saturated, there is an h ∈ L^{0}(Ω, X) such that law (f, g, h, G) = π. Then law (h) = αμ+(1−α)ν. It follows from τ(H) = law (h, G) = 1 that h is a measurable selection of G as required.

P 3: Suppose μ_{n} ∈ D_{F} for each n ∈ ℕ, and μ_{n} converges weakly to some μ ∈ M(X). Choose a measurable selection f_{n} of F such that law (f_{n}) = μ_{n}. Let G be the new correspondence from Ω to X such that for each ω, G(ω) is the closure of {f_{n}(ω) : n ∈ ℕ}. By Theorem III.30 in [2], G is a closed valued measurable correspondence. Since law (f_{n}) converges weakly to μ, Lemma 2.1 implies that the sequence law (f_{n}, G) has a subsequence that converges weakly to some measure ν such that marg_{X}(ν) = μ. Without loss of generality, we assume that the sequence law (f_{n}, G) converges weakly to ν. Hence, ν(H) ≥ lim sup_{n→∞} law (f_{n}, G)(H) = 1, which implies that μ(H) = 1. Because (Ω, A, P) is saturated, there exists f ∈ L^{0}(Ω, X) such that law (f, G) = ν. It follows that law (f) = μ, and d(f(ω), G(ω)) = 0 P -almost surely. Since F is closed valued, we have G(ω) ⊆ F (ω) for each ω. Therefore f is a measurable selection of F. Thus μ = law (f) ∈ D_{F}, and hence D_{F} is closed.

P 4: This follows from P 3 and the first two paragraphs of the proof of Theorem 4 in [24].

P 5: We show that every atomless probability space that satisfies P 4 satisfies P 5. By P 4, D_{F} is compact. By the assumptions that G is closed valued and F dominates G_{y} for each y ∈ Y, we know that H(y) = D_{G_{y}} is a closed subset of D_{F}. Therefore H is a compact valued correspondence from Y to M(X). To show that H is upper semi-continuous, let y_{n} converge to y ∈ Y, and for each n let μ_{n} ∈ H(y_{n}) = D_{G_{yn}} and let f_{n} be a measurable selection of G_{y_{n}} such that law (f_{n}) = μ_{n} with lim_{n→∞} μ_{n} = μ. Define a new correspondence J from (Ω, A, P) to X × Y by

J(ω) = cl{(f_{n}(ω), y_{n}) : n ∈ ℕ}.

By Theorem III.30 in [2], J is a closed valued measurable correspondence. Let Y_{0} be the compact set {y} ∪ {y_{n} : n ∈ ℕ}. Since J(ω) ⊆ F (ω) × Y_{0} for each ω, J is compact valued. By P 4, D_{J} is compact.

It is clear that law (f_{n}, y_{n}) converges weakly to μ ⊗ δ_{y}, where δ_{y} is the Dirac measure at y. Since law (f_{n}, y_{n}) belongs to D_{J} and D_{J} is compact, we know that μ⊗δ_{y} ∈ D_{J}. Hence, there is a measurable selection (f, y) of J such that law (f, y) = μ ⊗ δ_{y}. By (c), μ ∈ D_{G_{y}} = H(y). By Fact 3.4, H is upper semi-continuous.

P 6: Since G is measurable, it follows that the function ω ↦ G(ω)(B) is measurable for each Borel set B in X. Let F be the correspondence from Ω to X such that F (ω) = supp G(ω). It is easily checked that F is a closed valued, measurable correspondence. Note that G(ω)(F (ω)) = 1 for all ω. Let μ be the probability measure on X such that μ(B) = ∫_{Ω} G(ω)(B) dP for each Borel set B in X. We must find an f ∈ L^{0}(Ω, X) such that law (f) = μ, which gives (a), and f is a measurable selection of F, which gives (b).

For every open subset O of X, we have

F^{−1}(O) = {ω : F (ω) ∩ O ≠ ∅} = {ω : G(ω)(O) > 0}.

Thus F^{−1}(O) is measurable, and

μ(O) = ∫_{Ω} G(ω)(O) dP = ∫_{G(ω)(O)>0} G(ω)(O) dP
≤ P (G(ω)(O) > 0) = P (F^{−1}(O)).

So for every open set O in X we have

μ(O) ≤ P (F^{−1}(O)).

Then by Proposition 3.5, μ belongs to the closure of D_{F}. By P 3, D_{F} is closed, so μ ∈ D_{F} and there is a measurable selection f of F such that law (f) = μ.

3.3. Global Converse Results. In [24] it was shown that each of the properties P 1–P 6 in Theorem 3.6 fails for the Lebesgue unit interval (T, L, λ). That paper gave a correspondence G from (T, L, λ) to [−1, 1] that is a counterexample to each of P 1–P 4, and also gave counterexamples to P 5 and P 6 on the space (T, L, λ). In the following, we adapt these counterexamples to show that each part of Theorem 3.6 fails for every non-saturated atomless probability space.

Theorem 3.7. Each of the properties P 1–P 6 in Theorem 3.6 fails for every atomless probability space that is not saturated.

Proof. Let T = [0, 1], let (T, L, λ) be the Lebesgue unit interval, and let G be the correspondence from T to the closed interval [−1, 1] such that G(t) = {t, −t} for all t ∈ T. Then, D_{G} is neither closed nor convex, and the uniform distribution μ on [−1, 1] is not in D_{G} (see Example 1 of [24]). There is another correspondence G′ on (T, L, λ) such that law (G) = law (G′) but μ ∈ D_{G′}, so D(G) is not maximal. Moreover, there is a sequence of finite valued correspondences G_{n} such that μ ∈ D_{G_{n}} and G_{n}(t) → G(t) for each t, so property P 5 fails for (T, L, λ) (see Example 3 of [24]).

Suppose that (Ω, A, P) is an atomless probability space that is not saturated. By Fact 2.5 there is a set S ∈ A such that P (S) > 0 and A^{S} is countably generated (modulo sets of measure 0). Let P^{S} be the probability measure on (S, A^{S}) rescaled from P. As in the proof of Theorem 2.7, there is a measurable mapping h from S to T such that h induces an isomorphism between the corresponding measure algebras of (S, A^{S}, P^{S}) and the Lebesgue unit interval (T, L, λ). Thus, one can convert a counterexample on the unit Lebesgue interval to a counterexample on the non-saturated probability space (Ω, A, P) through h.

Let B denote the Borel subsets of [−1, 1]. Let F be the correspondence from Ω to [−1, 1] defined by setting F (ω) = G(h(ω)) for ω ∈ S and F (ω) = {0} for ω ∉ S. For any measurable selection f of F, f(ω) = 0 for ω ∉ S, and there is a Borel measurable mapping φ from T to [−1, 1] such that f(ω) = φ(h(ω)) for ω ∈ S. It is clear that φ is a measurable selection of G. Let δ_{0} be the probability measure on [−1, 1] such that δ_{0}({0}) = 1. It is then straightforward to check that D_{F} = (1−P (S)){δ_{0}}+P (S)D_{G}. Hence, D_{F} is neither maximal, closed, nor convex. Therefore, properties P 1–P 4 fail for the non-saturated probability space (Ω, A, P). Moreover, putting F_{n}(ω) = G_{n}(h(ω)) on S and F_{n}(ω) = 0 for ω ∉ S, we see that property P 5 also fails for (Ω, A, P).

Let H be the A-measurable mapping from Ω to M([−1, 1]) defined by setting H(ω) = (δ_{h(ω)} + δ_{−h(ω)})/2 for ω ∈ S and H(ω) = δ_{0} for ω ∉ S. Suppose that there is an A-measurable mapping f from Ω to [−1, 1] such that f is a measurable selection of F (which is supp H), and for every Borel set B in [−1, 1], law (f)(B) = ∫_{Ω} H(ω)(B)dP. Thus, law (f) = (1 − P (S))δ_{0} + P (S)μ, which implies that μ ∈ D_{G}. This is a contradiction. Hence property P 6 fails for the non-saturated probability space (Ω, A, P).

3.4. Local Converse Results. We will use Theorem 2.7 on local saturation to prove local converses for each of the parts P 1–P 4 and P 6 of Theorem 3.6. These improve the global converse results in Theorem 3.7, and are proved by a different method. We do not have a nice local converse for P 5, because P 5 involves an infinite family of correspondences. In this section it will always be understood that F, G are closed valued measurable correspondences to a Polish space X. For properties P 1–P 4, we will focus on the set of correspondences F that have a particular law ν ∈ M(F_{X}).

To get local converse results for a particular law ν, we need ν to be sufficiently powerful—we need ν to be atomless, and to be traceable as in Definition 3.8 below. Traceable correspondences are powerful in the sense that the original correspondence can be recovered from any selection. Our local converse result will show that (Ω, A, P) must already be saturated if there is just one atomless traceable measure ν and one of the properties P 1–P 4 that holds for all F ∈ law^{−1}(ν) on (Ω, A, P) but fails for some G ∈ law^{−1}(ν) on the Lebesgue unit interval (T, L, λ).

Definition 3.8. A Borel set S ⊆ F_{X} is traceable if there is a Borel function ψ : X → F_{X} such that ψ(x) = U whenever x ∈ U ∈ S. A correspondence F is traceable if there is a traceable set S such that F (ω) ∈ S a.s. We also say that F is traced by ψ.

It is easy to see that if F is traced by ψ and law (F) = law (G), then G is traced by ψ. So if F is traced by ψ and ν = law (F), we may say without ambiguity that ν is traceable, and that ν is traced by ψ.

Example 3.9. (i) The correspondence G from the Lebesgue unit interval (T, L, λ) to [−1, 1] used in the proof of Theorem 3.7 is traced by the Borel function ψ(s) = {s, −s}. Recall that each of the properties P 1–P 4 fail for G. We also note that the correspondence G is such that σ(G) = L, and that law (G) is atomless.

(ii) For any G, we can build a new correspondence Ĝ to F_{X} × X that “carries along” G, by defining Ĝ(·) = {G(·)} × G(·). Ĝ is closed-valued and measurable, and is traced by the Borel function ψ(U, x) = {U} × U. Note that if any one of the properties P 1–P 4 fails for G, then that property also fails for Ĝ.

Part (ii) of the example shows that every correspondence can be upgraded in a canonical way to a traceable correspondence. Here are our local converses for P 1–P 4. Example 3.9 shows that these results are not vacuous.

Theorem 3.10. Let ν ∈ M(F_{X}) be atomless and traceable. Then each of the following holds.

P 1(ν): Suppose that (Ω, A, P) has the property that D_{F} is maximal whenever law (F) = ν, but the Lebesgue unit interval (T, L, λ) does not have this property. Then (Ω, A, P) is saturated.

P 2(ν): Suppose that (Ω, A, P) has the property that D_{F} is convex whenever law (F) = ν, but this property fails for (T, L, λ) with a counterexample G such that σ(G) = L. Then (Ω, A, P) is saturated.

P 3(ν): Suppose that (Ω, A, P) has the property that D_{F} is closed whenever law (F) = ν, but (T, L, λ) does not have this property. Then (Ω, A, P) is saturated.

P 4(ν): Assume that ν{Z ∈ F_{X} : Z compact} = 1. Suppose (Ω, A, P) has the property that D_{F} is compact whenever law (F) = ν, but (T, L, λ) does not have this property. Then (Ω, A, P) is saturated.

Before beginning the proof, we make some observations about traceable correspondences.

Suppose ψ : X → Y is Borel. For any random element f of X, let f_{ψ}(ω) = (ψ(f(ω)), f(ω)). Note that if law (f) = law (g) then law (f_{ψ}) = law (g_{ψ}), so if μ = law (f) we may unambiguously define μ_{ψ} = law (f_{ψ}).

If ν is traced by ψ, then whenever law (F) = ν and f is a measurable selection of F with law (f) = μ, we have f_{ψ}(ω) = (F (ω), f(ω)) a.s. and hence law (F, f) = μ_{ψ} and marg_{F_{X}} μ_{ψ} = ν.

Lemma 3.11. Suppose that ν is traced by ψ, and that μ ∈ D_{H} for some H such that law (H) = ν. Then (Ω, A, P) has the saturation property for μ_{ψ} if and only if for every correspondence F ∈ law^{−1}(ν) on (Ω, A, P), we have μ ∈ D_{F}.

Proof. Suppose that for every correspondence F ∈ law^{−1}(ν) on (Ω, A, P), we have μ ∈ D_{F}. Then every F ∈ law^{−1}(ν) has a measurable selection f with law (f) = μ, and so law (F, f) = μ_{ψ}. This shows that (Ω, A, P) has the saturation property for μ_{ψ}.

Now suppose (Ω, A, P) has the saturation property for μ_{ψ} and let F ∈ law^{−1}(ν). By hypothesis, there is a H ∈ law^{−1}(ν) that has a measurable selection h such that law (h) = μ. Then law (H, h) = μ_{ψ}. By the saturation property for μ_{ψ}, there exists f such that law (F, f) = μ_{ψ}. Then law (f) = μ. Since h is a measurable selection of H, it follows from the proof of P1 in Theorem 3.6 that f is a measurable selection of F, and hence μ ∈ D_{F}.

We now prove Theorem 3.10.

Proof. In this proof we will deal with the two probability spaces (Ω, A, P) and the Lebesgue unit interval (T, L, λ). When we use the law^{−1} notation, it will always be understood to be with respect to (Ω, A, P).

By Theorem 3.6, there is an H on a saturated probability space such that law (H) = ν and D_{H} is maximal. We let D = D_{H}. Then for any F with law (F) = ν, D_{F} is maximal if and only if D_{F} = D. Moreover, by Theorem 3.6, D is convex and closed.

By hypothesis, ν is traced by the Borel function ψ : X → F_{X}. Note that since ν is atomless, μ_{ψ} is atomless for every μ ∈ D.

P 1(ν): By hypothesis there is a correspondence G on (T, L, λ) such that law (G) = ν and D_{G} is not maximal, so there is a measure μ ∈ D \ D_{G}. Therefore by Lemma 3.11, (T, L, λ) does not have the saturation property for μ_{ψ}. But for each F ∈ law^{−1}(ν), D_{F} is maximal, so μ ∈ D_{F}. By Lemma 3.11, (Ω, A, P) does has the saturation property for μ_{ψ}. By Theorem 2.7, (Ω, A, P) is saturated.

P 2(ν): By hypothesis, there is a correspondence G on (T, L, λ), measures μ_{1}, μ_{2} ∈ D_{G}, and α ∈ (0, 1) such that law (G) = ν, σ(G) = L, and

μ = αμ_{1} + (1 − α)μ_{2} ∉ D_{G}.

Since D_{G} ⊆ D and D is convex, μ ∈ D. By Lemma 3.11, (T, L, λ) does not have the saturation property for μ_{ψ}. We have μ_{1} = law (g_{1}) and μ_{2} = law (g_{2}) for some measurable selections g_{1}, g_{2} of G. Since σ(G) = L, g_{1} and g_{2} are σ(G)-measurable, so there are Borel functions ψ_{1}, ψ_{2} such that g_{1} = ψ_{1}(G), g_{2} = ψ_{2}(G).

It follows that for each F ∈ law^{−1}(ν), μ_{1} = law (ψ_{1}(F)) and μ_{2} = law (ψ_{2}(F)) belong to D_{F}. By hypothesis, D_{F} is convex, so μ belongs to D_{F}. By Lemma 3.11, (Ω, A, P) has the saturation property for μ_{ψ}. By Theorem 2.7, (Ω, A, P) is saturated.

P 3(ν): By hypothesis, there is a correspondence G on (T, L, λ) such that law (G) = ν and D_{G} is not closed. Fix μ in cl(D_{G}) \ D_{G}. Since D_{G} ⊆ D and D is closed, μ ∈ D. By Lemma 3.11, (T, L, λ) does not have the saturation property for μ_{ψ}.

Take any F ∈ law^{−1}(ν) on (Ω, A, P). By Proposition 3.5, for every open set O in X we have μ(O) ≤ λ(G^{−1}(O)) = P (F^{−1}(O)), and hence μ ∈ cl(D_{F}). Since D_{F} is closed, μ ∈ D_{F}. By Lemma 3.11, (Ω, A, P) has the saturation property for μ_{ψ}. By Theorem 2.7, (Ω, A, P) is saturated.

P 4(ν): By hypothesis, there is a compact valued correspondence G on (T, L, λ) such that law (G) = ν and D_{G} is not compact. By Theorem 3.6, D_{G} is contained in a compact set. Therefore D_{G} is not closed. The result now follows from P 3(ν).

To get a local converse to P 6, we need a notion of a traceable measurable mapping from Ω to M(X) that is analogous to the notion for correspondences.

Definition 3.12. A Borel set S ⊆ M(X) is traceable if there is a Borel function ψ : X → M(X) such that ψ(x) = τ whenever τ ∈ S and x ∈ supp τ. A measurable mapping G from Ω to M(X) is traceable if there is a traceable set S such that G(ω) ∈ S a.s. We also say that G is traced by ψ.

Note that if G is traced by ψ and law (G′) = law (G), then G′ is traced by ψ. If G is traced by ψ and ν = law (G), we say that ν is traceable, and that ν is traced by ψ.

As in the case of correspondences, if ν is traced by ψ, law (G) = ν, F (ω) = supp G(ω), and f is a measurable selection of F with μ = law (f), then f_{ψ}(ω) = (G(ω), f(ω)) a.s., law (G, f) = μ_{ψ}, and marg_{M(X)} μ_{ψ} = ν.

Example 3.13. (i) The measurable mapping G(t) = (δ_{t} + δ_{−t})/2 on the Lebesgue unit interval (T, L, λ) is traceable. This mapping was used in the proof of Theorem 3.7 above, and it was shown in [24] that property P 6 fails for G. We also note that law (G) is atomless.

(ii) Let G be a measurable mapping from (Ω, A, P) to M(X). Let Ĝ be the measurable mapping from (Ω, A, P) to M(M(X) × X) such that for each ω ∈ Ω, Ĝ(ω) = δ_{G(ω)} ⊗ G(ω) where δ_{y} is the Dirac measure at y and ⊗ is the independent product operation. Note that

supp Ĝ(ω) = {G(ω)} × supp G(ω).

Therefore Ĝ is traced by the Borel function ψ(τ, x) = δ_{τ} ⊗ τ. We note that if property P 6 fails for G, then it fails for Ĝ.

Here is our local converse result for P 6. Again, Example 3.13 shows that the result is not vacuous.

Theorem 3.14. Let ν ∈ M(M(X)) be atomless and traceable. Suppose that (Ω, A, P) has the property that P 6 holds for G whenever law (G) = ν, but the Lebesgue unit interval (T, L, λ) does not have this property. Then (Ω, A, P) is saturated.

Proof. Let ν be traced by the Borel function ψ : X → M(X). There is a unique measure μ ∈ M(X) depending only on ν such that whenever law (G) = ν on some probability space (Γ, C, Q) and B is Borel in M(X) × X,

μ(B) = ∫_{Γ} G(γ)(B) dQ.

Since ν is traced by ψ, marg_{M(X)} μ_{ψ} = ν.

Let G ∈ law^{−1}(ν) on (Ω, A, P). By hypothesis, there is a random element f of X that satisfies conditions (a) and (b) of P 6. By (a), law (f) = μ. By (b), f is a measurable selection of F where F (ω) = supp G(ω). Since ν is traced by ψ, law (G, f) = μ_{ψ}. This shows that (Ω, A, P) has the saturation property for μ_{ψ}.

By hypothesis, on the Lebesgue unit interval (T, L, λ) there is a random element G′ of M(X) such that law (G′) = ν but P 6 fails for G′. We will prove that there is no h with law (G′, h) = μ_{ψ}. This will show that (T, L, λ) does not have the saturation property for μ_{ψ}, and by Theorem 2.7 it will follow that (Ω, A, P) is saturated. Assume to the contrary that there is an h with law (G′, h) = μ_{ψ}. Then law (h) = μ, so (G′, h) satisfies condition (a). Let f be the random element on (Ω, A, P) from the preceding paragraph, with law (G, f) = μ_{ψ}. Since law (G′, h) = μ_{ψ} = law (G, f) and f is a measurable selection of supp G(·), h is a measurable selection of supp G′(·). Thus (G′, h) satisfies condition (b) as well, contradicting our hypothesis that property P 6 fails for G′. This proves that there is no h with law (G′, h) = μ_{ψ}, as required.

4. Large Games on Saturated Probability Spaces

In this section we further develop our underlying themes in the context of games. We show that saturated spaces have the desirable property that every game with a large number of players has a Nash equilibrium, and that no other probability spaces have this property. We also show that saturated spaces have the property that the set of laws of Nash equilibria for each game is closed in the weak topology, and we use Theorem 2.9 to get a strong local converse for that fact.

We shall first give a formal definition of a game based on a probability space of players (Ω, A, P). Let A be a compact metric space, and let U_{A} be the space of realvalued continuous functions on A × M(A) endowed with the sup-norm topology. By a game G with player space Ω and action space A we will mean a random element of U_{A} on (Ω, A, P). Thus, a game simply associates each player ω ∈ Ω with a payoff function G(ω)(a, τ) that depends on the player’s own action a and the distribution τ of actions by all the players. To improve readability, we also use G_{ω} to denote G(ω).

We will concentrate on the case that (Ω, A, P) is atomless. We say that such a game is “large” because it has at least continuum many players.

Definition 4.1. A Nash equilibrium of a game G is a random element g of A such that for P -almost all ω ∈ Ω,

G_{ω}(g(ω), law (g)) ≥ G_{ω}(a, law (g))

for all a ∈ A.

Thus, if g is a Nash equilibrium, then the distribution of actions by all the players is law (g) and every player chooses her optimal action g(ω) under this societal distribution. Note that we only consider pure-strategy Nash equilibria here.

Mas-Colell [20] introduced a corresponding notion of a measure game and Nash equilibrium distribution. A measure game with action space A is a probability measure ν ∈ M(U_{A}).

Definition 4.2. A Nash equilibrium distribution of a measure game ν is a probability measure μ ∈ M(U_{A} × A) such that marg_{U_{A}} μ = ν, and

μ{(u, x) : (∀a ∈ A)u(x, marg_{A} μ) ≥ u(a, marg_{A} μ)} = 1.

It is easy to see that for any game G with action space A and any random element g of A, g is a Nash equilibrium of G if and only if law (G, g) is a Nash equilibrium distribution of law (G). Given a Nash equilibrium distribution μ of a measure game ν, we say that a probability space (Ω, A, P) realizes μ if every game G ∈ law^{−1}(ν) on that space has a Nash equilibrium g such that law (G, g) = μ.

As a special case of Theorem 2.7, we immediately get a local characterization of saturated spaces in terms of Nash equilibria.

Corollary 4.3. Let ν be a measure game that is atomless as a measure, and μ be a Nash equilibrium distribution for ν. Suppose the Lebesgue unit interval (T, L, λ) does not realize μ. Then an atomless probability space (Ω, A, P) is saturated if and only if it realizes μ.

Proof. We see from the definitions that a probability space realizes a Nash equilibrium distribution μ if and only if it has the saturation property for μ.

Corollary 4.4. Let A be a compact metric space and let ν be an atomless measure on U_{A} that has a unique Nash equilibrium distribution. Suppose that (Ω, A, P) has the property that every G ∈ law^{−1}(ν) has a Nash equilibrium, but the Lebesgue unit interval (T, L, λ) does not. Then (Ω, A, P) is saturated.

Proof. Let μ be the unique Nash equilibrium distribution of ν. Then μ is atomless. Since μ is unique, whenever law (G) = ν and g is a Nash equilibrium of G we have law (G, g) = μ. It follows that (Ω, A, P) realizes μ but (T, L, λ) does not. By Corollary 4.3, (Ω, A, P) is saturated.

We now turn to the question of the existence of Nash equilibria. We will use Mas-Colell’s existence theorem for Nash equilibrium distributions.

Fact 4.5. (Mas-Colell [20]). Every measure game with compact metric action space A has a Nash equilibrium distribution.

Our next theorem gives a global characterization of saturated probability spaces by the existence of Nash equilibria.

Theorem 4.6. Let (Ω, A, P) be an atomless probability space, and A an uncountable compact metric space. Then (Ω, A, P) is saturated if and only if every game G with player space (Ω, A, P) and action space A has a Nash equilibrium.

Proof. Suppose that (Ω, A, P) is saturated. By Fact 4.5, the measure game ν = law (G) has a Nash equilibrium distribution μ. By full saturation, there is a random element g of A such that law (G, g) = μ. Therefore g is a Nash equilibrium of G.

Instead of using Fact 4.5, one can also get a Nash equilibrium of G by using the proof of Theorem 1 of [15]. One only needs to use the distributional properties of correspondences on saturated probability spaces (instead of Loeb spaces) such as the convexity, compactness, and preservation of upper semi-continuity.^{2}

For the converse, we first consider the case that the action space A is the interval [−1, 1]. By Corollary 4.4 it suffices to find a game G on the Lebesgue unit interval (T, L, λ) such that G has no Nash equilibrium and law (G) is atomless and has a unique Nash equilibrium distribution. Let G be the game defined in Section 2 of [14], and Example 3 in [23], where G is a one-to-one continuous mapping from T to U_{A}. Let U be the image of G, which is a compact set in U_{A}. It is shown in [14] that G has no Nash equilibrium. It is clear from the definition in [14] that ν = law (G) is atomless. The uniqueness of the Nash equilibrium distribution for the measure game law (G) is implicit in the proof in [14] and [23], and can be seen as follows.

Let μ be any Nash equilibrium distribution for the measure ν. Let G′ be a game on (Ω, A, P) with a Nash equilibrium f and law (G′, f) = μ. By modifying the definition of G′ on a null set, we may assume that G′ takes values in the set U. As shown in [14] and [23], law (f) is the uniform distribution on [−1, 1], and when G′(ω) = G(t), the best response is t or −t. Fix any ω ∈ Ω such that f(ω) is a best response; note that such elements ω ∈ Ω form a set of P -full measure. Then, there is a unique t ∈ T such that G′(ω) = G(t), which implies that f(ω) must be t or −t. Hence, we have G′(ω) = G(|f(ω)|). Therefore, μ = law (G′, f) = law (G(|f(·)|), f) is the unique Nash equilibrium distribution for ν.

Finally, we prove the converse for the general case that A is any fixed uncountable compact metric space. Suppose that (Ω, A, P) is not saturated. By the converse for the case that [−1, 1] is the action space, there is a game G^{1} with player space (Ω, A, P) and action space [−1, 1] but without a Nash equilibrium.

2In fact, the full statements of Theorem 1 for large games and Theorem 3 for finite-player games with incomplete information in [15] can be restated on saturated probability spaces instead of Loeb spaces; exactly the same proofs work in this more general situation. The same thing also works for Theorem 2 of [15] by using integration of correspondences on saturated probability spaces.

It is noted in [23], page 339, that there exists a continuous surjective mapping F from A to [−1, 1] and a continuous injective mapping F^{0} from U_{[−1,1]} to U_{A} such that F^{0}(u)(x, y) = u(F (x), yF^{−1}) whenever u ∈ U_{A}, x ∈ A, y ∈ M(A). We can now define a new game G^{2} with player space (Ω, A, P) and action space A by using the composition mapping G^{2} = F^{0} ∘ G^{1}. Suppose f^{2} is a Nash equilibrium for the new game G^{2}. Then it can be easily checked that the composition mapping f^{1} = F ∘ f^{2} is a Nash equilibrium for the game G^{1}. This is a contradiction. Hence the set of Nash equilibria of G^{2} is empty, and the converse for the general case is shown.

For each game G, let

E_{G} = {law (G, g) : g is a Nash equilibrium of G}.

Note that every measure μ ∈ E_{G} is a Nash equilibrium distribution of law (G). We now prove that saturated probability spaces have a closure property for Nash equilibria, and then use Theorem 2.9 to get a local converse.

Theorem 4.7. Suppose (Ω, A, P) is saturated. Then for every game G on Ω with compact metric action space A, the set E_{G} is closed in M(U_{A} × A).

Proof. Suppose μ_{n} ∈ E_{G} and μ_{n} converges weakly to μ. Take a Nash equilibrium g_{n} of G such that law (G, g_{n}) = μ_{n}. By full saturation, there is a g ∈ L^{0}(Ω, A) with law (G, g) = μ. Let ν_{n} = law (g_{n}) and ν = law (g). Then ν_{n} converges weakly to ν. Since A is separable it has a countable dense subset A_{0}.

Fix any a ∈ A_{0}. We have

G_{ω}(g_{n}(ω), ν_{n}) ≥ G_{ω}(a, ν_{n}) a.s.

By [1], Theorem 4.4 on page 27, law (G, g_{n}, ν_{n}) converges weakly to law (G, g, ν) in M(U_{A} × A × M(A)). It follows that h_{n}(ω) = G_{ω}(g_{n}(ω), ν_{n}) − G_{ω}(a, ν_{n}) converges weakly to h(ω) = G_{ω}(g(ω), ν) − G_{ω}(a, ν). Since law (h_{n})([0, ∞)) = 1 for each n, we have law (h)([0, ∞)) ≥ lim sup_{n→∞} law (h_{n})([0, ∞)) = 1. Hence,

G_{ω}(g(ω), ν) ≥ G_{ω}(a, ν) a.s.

By grouping countably many null sets together, we see that for P -almost all ω ∈ Ω,

(∀a ∈ A) G_{ω}(g(ω), ν) ≥ G_{ω}(a, ν),

which means that g is a Nash equilibrium of G, so μ = law (G, g) ∈ E_{G}.

Corollary 4.8. For each measure game ν with compact metric action space A, the set of all Nash equilibrium distributions for ν is closed in M(U_{A} × A).

Here is our local converse for Theorem 4.7.

Corollary 4.9. Let ν be a measure game with compact metric action set A. Suppose that on (Ω, A, P), E_{G} is closed for every game G ∈ law^{−1}(ν), but on the Lebesgue unit interval (T, L, λ) there is a game G′ ∈ law^{−1}(ν) such that σ(G′) = L and E_{G′} is not closed. Then (Ω, A, P) is saturated.

Proof. Let C be the set of Nash equilibrium distributions for ν. For each game G with law (G) = ν, E_{G} = C(G). C is non-empty by Fact 4.5, so ν = marg_{U_{A}} μ for some μ ∈ C. Then by Theorem 2.9, (Ω, A, P) is saturated.

The following simple example shows that Corollary 4.9 is not vacuous.

Example 4.10. Let G′ be the game on the Lebesgue unit interval (T, L, λ) with action space A = [−1, 1] and the payoff function G′_{t}(a, τ) = −|t−|a||. Then σ(G′) = L. There is a game G with law (G) = law (G′) that has a Nash equilibrium g such that law (g) is the uniform probability measure on A. Then law (G, g) belongs to the closure of E_{G′} but not to E_{G′}.

5. From One Saturated Probability Space to Another

In this section, we show that if the properties P 1–P 6 in Theorem 3.6 are established for one saturated probability space, then they hold for every other saturated probability space. The key is the choice of appropriate mappings for applying full saturation. This result combined with the known theorems in [24] that properties P 1–P 6 hold for atomless Loeb probability spaces, and the fact that atomless Loeb probability spaces are saturated, gives an alternative proof of Theorem 3.6. We will also prove analogous results for Theorems 4.6 and 4.7 concerning Nash equilibria in large games.

Of course, each of the properties under discussion was already proved outright for all saturated probability spaces in Theorem 3.6 and in Section 4. Our point here is that if we add the hypothesis that the properties hold for some particular saturated probability space, then there is a very simple proof that the properties hold for all saturated probability spaces.

This demonstrates a general technique for extending certain types of results from atomless Loeb probability spaces (or even the simplest hyperfinite Loeb counting spaces) to all saturated probability spaces. Thus, hyperfinite Loeb counting spaces can play a prototype role in the class of all saturated probability spaces.

Lemma 5.1. Let (Ω, A, P) and (Γ, C, Q) be two saturated probability spaces. For any closed measurable correspondences F on (Ω, A, P) and F′ on (Γ, C, Q), if law (F) = law (F′) then D_{F} = D_{F′}.

Proof. Let μ ∈ D_{F}. We have μ = law (f) for some measurable selection f of F. Since (Γ, C, Q) is saturated, there is a random element f′ of X such that law (F′, f′) = law (F, f). Because f is a selection of F and law (F′, f′) = law (F, f), it follows from the proof of P1 in Theorem 3.6 that f′ is a selection of F′. Therefore law (f′) ∈ D_{F′}. This shows that D_{f′} ⊇ F_{F}. The other inclusion follows by symmetry.

The following proposition shows that if all the regularity properties P 1–P 6 for distribution of correspondences hold for one particular saturated probability space, then they hold for any other saturated probability space.

Proposition 5.2. Let (Ω, A, P) and (Γ, C, Q) be two saturated probability spaces. Assume that each of the properties P 1–P 6 hold for (Ω, A, P). Then each of the properties P 1–P 6 hold for (Γ, C, Q).

Proof. P 1: Let F′ be a closed valued measurable correspondence from (Γ, C, Q) to X. Since (Ω, A, P) is atomless, there is a measurable mapping F from (Ω, A, P) to F_{X} such that law (F) = law (F′). It follows that F is a closed valued, measurable correspondence from (Ω, A, P) to X. Lemma 5.1 shows that D_{F} = D_{F′}. By property P 1 for (Ω, A, P), D_{F} is maximal, so D_{F′} is also maximal.

P 2: The convexity of the set of laws of measurable selections of an arbitrary correspondence follows from the case of a correspondence consisting of two measurable functions. Without loss of generality, assume that F′ is a closed valued measurable correspondence on (Γ, C, Q).

As in the proof for P 1, there exists a closed valued, measurable correspondence from (Ω, A, P) to X such that law (F) = law (F′). By Lemma 5.1, D_{F′} = D_{F}. Since D_{F} is convex, so is D_{F′}.

P 3: We can assume without loss of generality that F′ is a closed valued measurable correspondence on (Γ, C, Q). The second paragraph in the proof of P 2 above shows that D_{F′} = D_{F} for some closed valued, measurable correspondence from (Ω, A, P) to X. Since D_{F} is closed, so is D_{F′}.

P 4: Let C_{X} be the space of compact subsets of X endowed with the Hausdorff metric, which is a complete separable metric. We can assume without loss of generality that F′ is a compact valued measurable correspondence on (Γ, C, Q). The second paragraph in the proof of P 2 shows that D_{F′} = D_{F} for some closed valued, measurable correspondence F from (Ω, A, P) to X with law (F) = law (F′). Since F′ is compact valued, law (F′)(C_{X}) = 1, and hence law (F)(C_{X}) = 1. This means that one can take F to be compact valued. Since D_{F} is compact, so is D_{F′}.

P 5: Since property P 4 holds for (Ω, A, P), the preceding paragraph shows that P 4 holds for (Γ, C, Q). The proof of Theorem 3.6 shows that any atomless probability space that has property P 4 has property P 5. Thus P 5 holds for (Γ, C, Q).

P 6: Let G′ be a measurable mapping from (Γ, C, Q) to the space M(X) of probability measures on X, and F′ a correspondence from (Γ, C, Q) to X such that F′(γ) = supp G′(γ) for each γ ∈ Γ.

As in the proof of P 2, let G and F be measurable mappings from (Ω, A, P) to M(X) and F_{X} respectively such that law (G, F) = law (G′, F′). It follows that F is a closed valued, measurable correspondence from (Ω, A, P) to X. Since G′(γ)(F′(γ)) = 1 for all γ ∈ Γ, we know that G(ω)(F (ω)) = 1 for almost all ω.

There is a measurable mapping f from (Ω, A, P) to X such that (i) for every Borel set B in X, law (f)(B) = ∫_{Ω} G(ω)(B)dP; (ii) for each ω ∈ Ω, f(ω) ∈ supp G(ω) ⊆ F (ω). By full saturation, there is a measurable mapping f′ from (Γ, C, Q) to X such that law (G, F, f) = law (G′, F′, f′). This f′ will have the desired property.

Note that for each of the properties P 1–P 4 and P 6, the above proof of that property for (Γ, C, Q) used only the assumption that the same property holds for (Ω, A, P); it did not use any of the results from Section 3. To prove that P 5 holds for (Γ, C, Q), we used the assumption that P 4 holds for (Ω, A, P) and a fact from Subsection 3.1.

The following proposition shows (without using Mas-Colell’s result stated in Fact 4.5) that if the existence result for large games as stated in Theorem 4.6 holds for one particular saturated probability space as the player space, then it holds for any other saturated probability space as the player space. A similar statement holds for the closure result in Theorem 4.7.

Proposition 5.3. Let (Ω, A, P) and (Γ, C, Q) be two saturated probability spaces, and let A be a compact metric space.

(i) If every game with player space (Ω, A, P) and compact metric action space A has a Nash equilibrium, then every game with player space (Γ, C, Q) and the same action space A has a Nash equilibrium.

(ii) If for every game with player space (Ω, A, P) and compact metric action space A, E_{G} is closed, then for every game G′ with player space (Γ, C, Q) and the same action space A, E_{G′} is closed.

Proof. (i) Suppose that every game G with player space (Ω, A, P) and action space A has a Nash equilibrium. Let G′ be a game with player space (Γ, C, Q) and the same action space A. By Lemma 2.1 (ii) there is a random element G of U_{A} on (Ω, A, P) such that law (G) = law (G′). The game G has a Nash equilibrium g ∈ L^{0}(Ω, A), so law (G, g) is a Nash equilibrium distribution for law (G). By full saturation there is a random element g′ of A on (Γ, C, Q) such that law (G′, g′) = law (G, g). Hence g′ is a Nash equilibrium for G′.

(ii) An argument like the proof of part (i) shows that for any games G on (Ω, A, P) and G′ on (Γ, C, Q) with law (G) = law (G′), we have E_{G} = E_{G′}. Since E_{G} is closed, it follows that E_{G′} is closed as well.

Acknowledgments. This work was initiated when Yeneng Sun visited the University of Wisconsin-Madison in May 2000. The first draft, [13], was written in April 2002. In the earlier version, the distributional properties of correspondences and the existence of pure strategy equilibria in large games on saturated probability spaces were obtained from the corresponding results on Loeb measure spaces via full saturation. The present version, which was completed in July 2008, gives simpler proofs for the results on correspondences and games using the full saturation directly, rather than using parallel results for Loeb spaces. The local converse results for correspondences and games are also new to this version. Some of our results have also been reported at various places, including a 2002 ICM satellite conference Symposium on Stochastics and Applications, Singapore, August 15-17, 2002 (http://ww1.math.nus.edu.sg/ssa/abstracts/YenengSunAbstract.PDF), and the Workshop on Mathematical Logic and its Applications, Singapore, June 17-18, 2004; seminar talks at the City University of Hong Kong in December 2003, the Academia Sinica, Taiwan in December 2005, the University of Illinois at Urbana-Champaign in October 2006. Some results as presented in the earlier draft [13] have also been used by colleagues in several later papers [18], [21], [22] and [25].

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Department of Mathematics, University of Wisconsin,, 480 Lincoln Drive, Madison WI 53706, U.S.A.

E-mail address: keisler@math.wisc.edu

Department of Mathematics, National University of Singapore,, 2 Science Drive 2, Singapore 117543, Republic of Singapore

E-mail address: matsuny@nus.edu.sg