Changing the order of $\lim$ and $\sup$

Given $\epsilon>0$, let $E_n=\{x\in X: f_n(x)\ge \hat f-\epsilon\}$ and $E^*=\{x\in X: f(x)\ge \hat f-\epsilon\}$. We know that the sets $E_n$ are nonempty and nested: $E_{n+1}\subset E_n$. We would like to show that $E^*$ is nonempty. Since $f_n$ decreases to $f^*$ pointwise, it follows that $E^*=\bigcap E_n$.

So the problem becomes: how do we show that a nested sequence of nonempty sets has nonempty intersection? I know three ways:

• $E_1$ has finite measure and the measures of $E_n$ are bounded from below by $c>0$.
• each $E_n$ is compact
• $X$ is complete, each $E_n$ is closed, and $\mathrm{diam}\, E_n\to 0$.

Your problem is equivalent to asking if $$\begin{equation}\inf_n \sup_x f_n(x) = \sup_x \inf_n f_n(x) \tag{*}\end{equation}.$$

This question is the subject of minimax theorems, e.g. Sion's minimax theorem. Minimax theorems are applied frequently in game theory.

Generally speaking, (*) is true if $n \mapsto f_n(x)$ is (quasi-)concave and $x \mapsto f_n(x)$ is (quasi-)convex. You will also need some sort of (semi-)continuity and a "nice" structure for the sets in the infimum and supremum (for instance $x \in \mathbb{R}$ and $f_n(x) = \frac1n |g(x)|$).