On Remark 1.12, page 33 of Villani's Topics in optimal transport we find the following problem. It should be easy but it's turning me mad.
Let $c:\mathbb{R}^n\times\mathbb{R}^n\rightarrow \mathbb{R}_+\cup\{+\infty\}$ be a measurable, bounded, lower semicontinuous function, then there exists $c_l$ an increasing family of bounded uniformly continuous functions converging pointwise to $c$. Let $\varphi:\mathbb{R}^n\rightarrow\mathbb{R}$ be a bounded function.
Define $$\varphi^c(y)=\inf_{x\in X}[c(x,y)-\varphi(x)],$$ and $$\psi_l(y)=\inf_{x\in X}[c_l(x,y)-\varphi(x)].$$ Then, it is easily seen that each $\psi_l$ is uniformly continuous, but how do we prove that $\psi_l$ converges pointwise to $\varphi^c$?
This is not true as stated. Define $$\varphi(x) = \begin{cases} 0 \quad &\text{ if } x=0 \\ 1 \quad &\text{ if } x\ne 0 \end{cases}$$ and $c(x,y) = \varphi(x)$. Clearly, $\varphi^c(y)=0$ for all $y$.
Suppose that $c_l$ is a continuous function such that $c_l\le c$. Take a sequence of nonzero numbers $x_k$ converging to $0$: then $$\lim_{k\to\infty}(c_l(x_k,y)-\varphi(x_k)) = c_l(0,y)-1 \le -1$$ and therefore $\psi_l(y)\le -1$ for all $y$.
The author needed this "exercise" to justify the measurability of $\varphi^c$... If I were in your place, I would trust the claim that it's measurable and keep on reading.