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 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?
Another question along the line is as follows: suppose $f_n(x,y)$ converges to $f(x,y)$ pointwise and the associated partial minimization $\phi(y) = \inf\{f(x,y):x\in X\}$ and $\phi_n(y) = \inf\{f_n(x,y):x\in B_n\}$ are measurable, where $B_n$ is the ball with radius $n$. What can we say about the quantities
$$ \int \phi(y) d\nu \quad and \quad \int \phi_n(y) d\nu $$ ?
Any suggestions are appreciated!
For any $\pi\in\Pi(\nu)$ $$\int_{X\times Y} f(x,y) \text{d}\pi(x,y)\geq\int_Y\phi(y)\text{d}\nu(x)$$ Hence $$\inf_{\pi\in\Pi(\nu)}\int_{X\times Y} f(x,y) \text{d}\pi(x,y)\geq\int_Y\phi(y)\text{d}\nu(x)$$
For the other inequality, you can choose some $\pi_\varepsilon$ which is supported on a subset $A\subset X\times Y$ for which $f(x,y)\leq \phi(y)+\varepsilon\ \forall (x,y)\in A$ hence $$\inf_{\pi\in\Pi(\nu)}\int_{X\times Y} f(x,y) \text{d}\pi(x,y)\leq\int_{X\times Y} f(x,y) \text{d}\pi_\varepsilon(x,y)\leq \int_Y\phi(y)\text{d}\nu(x)+\varepsilon$$ for arbitrary $\varepsilon>0$. I think you do need to use a measurable selection to construct such a measure.
For the second question, since $\phi\geq\limsup\phi_n$, if we also assume $f_n$ are dominated by some measurable function then $$\int\phi\geq\int\limsup\phi_n\geq \limsup\int\phi_n$$ with equality by the dominated convergence theorem whenever $\phi=\lim\phi_n$. But otherwise, the inequality may be strict.