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.