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?



optimal transport, 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 $$\int\_{X\times Y} f(x,y)d\mu(x,y)=\int\_X f\circ s(x)d\nu(x)=\int\_X \phi(x)d\nu(x).$$


