I've noticed many posts regarding conditions for which you can interchange limits and essential supremum so please forgive me if this is somewhat of a duplicate. I believe my problem may be phrased differently to many of these posts. Here is my question:
- Suppose that $\mathrm{ess \ sup}_{x\in X}|f_n(x)-f(x)|\to 0$, can we deduce from this that $|f_n(x)-f(x)|\to 0$ for each $x\in X$ or vice versa? Or would this require uniform convergence of the sequence $(f_n)$ in order to interchange limits?
[EDIT:] The comment indicated that the question was not answered. Suppose for each $x\in X$ we know that $\vert f_n(x)-f(x)\vert\to 0$ as $n\to\infty$. What can be said about $\mathrm{ess\,sup}\vert f-f_n\vert$? The following example gives an answer.
Take $X = (0,1)$, $f_n(x) = cx^n$ and $f(x) = 0$. Then $f_n(x) \to f(x)$ for each fixed $x$ and $n\to\infty$ but $\mathrm{ess\,sup}\vert f_n(x)-f(x)\vert = \vert c\vert$ for each $n$.
Can anything be said to the converse? That is, if $\mathrm{ess\,sup}\vert f-f_n\vert \to 0$ can we say anything about the behavior of $\vert f(x)-f_n(x)\vert$ for each $x\in X$.
The answer is that if $D = \{x\in X: \liminf_{n}\vert f_n(x) - f(x)\vert > 0\}$ (the set of points which do not converge) then $\mu(D) = 0$.