Changing the order of $\lim$ and $\sup$ Suppose that $f_n:X\to [0,1]$ where $X$ is some arbitrary set. Suppose that 
$$
  f_n(x)\geq f_{n+1}(x)
$$
for all $x\in X$ and all $n = 0,1,2,\dots$ so there exists $\lim_n f_n(x)$ point-wise, let's call it $f(x)$. 
Define $f^*_n:=\sup\limits_{x\in X}f_n(x)$, $f^*:=\sup\limits_{x\in X}f(x)$ and $\hat f:= \lim\limits_n f^*_n$. I wonder when $f^* = \hat f$, i.e.
$$
  \lim\limits_n \sup\limits_{x\in X}f_n(x) = \sup\limits_{x\in X}\lim\limits_n f_n(x).
$$
I was googling the topic, but strangely have not found any information, strangely because I expected it to be available as for changing the order of limits or of integration.
Some simple facts: $\hat f\geq f^*$ and the reverse is true at least when $f_n$ converges uniformly to $f$. This does not hold in general, e.g. when $f_n = 1_{[n,\infty)}$.
I would appreciate any other ideas that you can advise me. Also related to this. A similar question was asked here.
 A: 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 $\mathrm{diam}\, E_n\to 0$. 

A: 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 $f_n(x) = \frac1n |g(x)|$).
