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

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.

• I think you can extend the result to the case where $c_n$ is only l.s.c. and bounded from below. By Portmanteau theorem, \begin{align*} \liminf_n \int c_n d\pi_n - \int c_{N} d \pi^* &\ge \liminf_n\int (c_n - c_{N}) d\pi_n + \liminf_n \int c_{N} d(\pi_n- \pi^*) \\ &\ge \liminf_n\int (c_n - c_{N}) d\pi_n + 0\\ &\ge 0 \quad \text{because} \quad c_n \nearrow c. \end{align*} Commented Nov 1, 2022 at 2:37