Let $X = Y = \mathbb{R}^d$ and let $\nu$ be a probability measure on $\mathbb{R}^d$. Consider the collection of probability measure $\pi$ on $X\times Y$ such that $\pi$ has $y$-marginal $\nu$:

$$ \Pi(\nu) = \{\pi: \pi(X,dy) = \nu(dy)\}. $$

Let $f:X\times Y \mapsto \mathbb{R}$ be a measurable function (you can assume it is continuous and has nice growth condition) such that the partial minimization

$$ \phi(y) = \inf\{f(x,y):x\in X\} $$ is measurable. My question: is it true that

$$ \inf_{\pi \in \Pi(\nu)} \int f(x,y) d\pi = \int \phi(y) d\nu, $$ and if so, can it be proved without using any sort of measurable selection?


The answer is affirmative if $f$ is Borel-measurable. It is suffices to prove the equality for $f$ having finitely many values. Let $V=f(X\times Y)$ be the finite set of values of $f$. By the Borel-measurability of $f$, for every $v\in V$ the set $D_v=f^{-1}(v)$ is Borel and its projection $A_v=pr_X(D_v)$ onto $X$ is analytic.

By the Uniformization Theorem 18.1 (of Jankov and von Neumann) in the book "Classical Descriptive Set Theory" of Kechris, there exist a uniformizing function $s_v:A_v\to D_v$ which is $\sigma(\Sigma^1_1)$-measurable and $pr_X\circ s_v(x)=x$ for all $x\in A_v$. Consider the $\sigma(\Sigma_1^1)$-measurable function $s:X\to X\times Y$ defined by $s(x)=s_{v}(x)$ where $v\in V$ is the unique element of $V$ such that $x\in A_v\setminus \bigcup_{V\ni u<v}A_u$. It follows that $f\circ s=\phi$.

Consider the probability measure $\mu$ on $X\times Y$ defined by $\mu(B)=\nu(s^{-1}(B))$ for any Borel subset $B\subset X\times Y$. This measure is well-defined as $s^{-1}(B)$ belongs to the $\sigma$-algebra $\sigma(\Sigma^1_1)$ and hence is $\nu$-measurable by Lusin Theorem 29.7 (in Kechris' book). If follows that $$\int_{X\times Y} f(x,y)d\mu(x,y)=\int_X f\circ s(x)d\nu(x)=\int_X \phi(x)d\nu(x).$$

  • $\begingroup$ Thanks for your answer. I guess the Uniformization Theorem here is some analogy of the Von Neumann measurable selection theorem? Just wondering if there is a more elementary proof without using such results. $\endgroup$ – Ryan Feb 12 at 22:40
  • $\begingroup$ @Ryan Oh sorry! I did not read your question carefully. You asked about the proof WITHOUT selections and I red it as WITH selections. Yes, I think there is a proof without these selection results, instead one can use the known results on the surjectivity of the map $Pf:PX\to PY$, induced by a surjective map $f:X\to Y$ from a Polish space $X$ to a compact space $Y$. $\endgroup$ – Taras Banakh Feb 13 at 6:38
  • $\begingroup$ Hi Taras, thanks for you prompt respond! Could you please elaborate on your previous comment, or refer me to some known results that you mentioned? Thanks again! $\endgroup$ – Ryan Feb 13 at 18:47
  • $\begingroup$ @Ryan Using the surjectivity of the map $Ppr_X:P(D_v)\to P(A_v)$, we can choose a $\sigma$-additive Borel measure $\mu_v$ on the Borel subset $D_v$ of $X\times Y$ such that $Ppr_X(\mu_v)=\nu{\restriction}A_v\setminus \bigcup_{u<v}A_u$. Then $\mu=\sum_{v\in V}\mu_v$ is a measure withessing the required equality. $\endgroup$ – Taras Banakh Feb 13 at 21:09
  • $\begingroup$ @Ryan So the problem reduced to the proof of the surjectivity of the maps $Ppr_X$. But I do not know how to prove this theorem without selection theorems. Maybe for your purposes it suffices to refer to some known results on the surjectivity of maps between spaces of measures? $\endgroup$ – Taras Banakh Feb 13 at 21:11

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