# 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.

-
Sorry, forget my comments. I was trying to be helpful without properly thinking. –  Rudy the Reindeer Jul 12 '12 at 14:45
@MattN: done: fogotten –  Ilya Jul 12 '12 at 14:46
I think that convergence of minima/maxima is related to $\Gamma$-convergence. –  Siminore Jul 12 '12 at 14:46

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 $E^*=\bigcap E_n$.
• $E_1$ 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 $\mathrm{diam}\, E_n\to 0$.
Thank you, did you mean that $f^*(x)$ is $f(x)$? –  Ilya Jul 12 '12 at 16:45