If $c_n \nearrow c$ then $\lim_n \inf_{\pi \in \Pi(\mu, \nu)} \int c_n d\pi = \inf_{\pi \in \Pi(\mu, \nu)} \int c d\pi$

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Let $X,Y$ be Polish spaces and $c:X \times Y \to \mathbb [0, +\infty]$ lower semi-continuous. Let $(c_n)$ with $c_n:X \times Y \to \mathbb [0, \infty)$ be a non-decreasing sequence of bounded continuous functions such that $c_n \nearrow c$. We fix Borel probability measures (b.p.m.) $\mu \in \mathcal P(X)$ and $\nu \in \mathcal P(Y)$. Let $\Pi(\mu, \nu)$ be the collection of b.p.m.'s on $X \times Y$ with marginals $\mu$ on $X$ and $\nu$ on $Y$.

Then $$ \lim_n \inf_{\pi \in \Pi(\mu, \nu)} \int c_n d\pi =:A = B:= \inf_{\pi \in \Pi(\mu, \nu)} \int c d\pi. $$

Answer 1

We have \begin{align} A &= \sup_n \inf_{\pi \in \Pi(\mu, \nu)} \int c_n d\pi \\ &\le \inf_{\pi \in \Pi(\mu, \nu)} \sup_n \int c_n d\pi \\ &= \inf_{\pi \in \Pi(\mu, \nu)} \int c d\pi \\ &=B. \end{align}

Let's prove the converse. There is $$ \pi_n \in \operatorname{argmin}_{\pi \in \Pi(\mu, \nu)} \int c_n d \pi \quad \forall n \in \mathbb N. $$

Because $\Pi(\mu, \nu)$ is uniformly tight and weak* closed, so $\Pi(\mu, \nu)$ is weak* compact by Prokhorov theorem. Then there is $\pi^* \in \Pi(\mu, \nu)$ such that $\pi_n \to \pi^*$ weakly. First, we prove that $$ \lim_n\int c_n d\pi_n \ge \int c_{N} d \pi^* \quad \forall N \in \mathbb N. $$

Indeed, $$ \int c_n d\pi_n - \int c_{N} d \pi^* = \int (c_n - c_{N}) d\pi_n + \int c_{N} d(\pi_n- \pi^*). $$

Then \begin{align} \lim_n \int c_n d\pi_n - \int c_{N} d \pi^* &= \lim_n\int (c_n - c_{N}) d\pi_n + \lim_n \int c_{N} d(\pi_n- \pi^*) \\ &= \lim_n\int (c_n - c_{N}) d\pi_n \\ &\ge 0 \quad \text{because} \quad c_n \nearrow c. \end{align}

It follows that $$ A=\lim_n\int c_n d\pi_n \ge \lim_N \int c_{N} d \pi^* = \int c d\pi^* \ge B. $$

This completes the proof.