# Convergence in $L^\infty$ is nearly uniform convergence

Let $\{f_n\}$ be a sequence of functions in $L^\infty$. I want to prove that $\{f_n\}$ converges to $f\in L^\infty$ $\Leftrightarrow$ there is a set $E$ of measure zero such that $f_n$ converges uniformly to $f$ on $E^c$.

My Attempt:
$(\Rightarrow)$ Suppose $f_n\rightarrow f$ in $L^\infty$. Then $\Vert f_n-f\Vert_\infty=\inf \{M:|f_n-f|\leqslant M ~~\text{a.e.}\}\lt \varepsilon$ for any $\varepsilon \gt 0$. Let $E=\{x:|f_n(x)-f(x)|\gt \varepsilon\}.$ The $m(E)=0$. So $f_n\rightarrow f$ uniformly on $E^c$.

$(\Leftarrow)$ Suppose $f_n\rightarrow f$ uniformly on $E^c$ with $m(E)=0$. Then $\forall~t\in E^c$ and $n\gt N$, $|f_n(t)-f(t)|\lt \varepsilon$. But
$$|f(t)|=|f(t)-f_n(t)+f_n(t)|\leqslant |f(t)-f_n(t)|+|f_n(t)|\lt \varepsilon + \Vert f_n(t)\Vert_\infty.$$ So $f$ is bounded and hence $f\in L^\infty$.

Please, could someone look over what I've done and point out any errors?
Thanks.

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The first direction is good. For the second direction, just a little change to make it seem more complete. It looks like your bound on $|f(t)|$ still depends on $n$ and that you forget to take a limit as $n\to\infty$ or something like that. Since your inequality holds for all $n> N$, say that specifically, we have $|f(t) | < \epsilon + \|f_{N+1} (t) \|_{\infty}.$ – Ragib Zaman Nov 19 '11 at 0:47
Thanks Ragib.${}{}{}{}$ – Colin Nov 19 '11 at 2:54

For the forward direction, unwinding the definitions, you need to show: there exists a set $E$ with $m(E)=0$ such that for every $\epsilon > 0$, there exists $N$ such that for all $n > N$ and all $x \in E^c$, we have $|f_n(x) - f(x)| < \epsilon$. Your set $E$ depends on $n$, so this is not good. Remember that "$f_n \to f$" is a statement about the sequence $\{f_n\}$, not a statement about $f_1$ or $f_6$ or any other particular function from the sequence.
So start with what you know: $||f_n - f||_{L^\infty} \to 0$. The quantity $||f_n - f||_{L^\infty}$ bounds $|f_n(x) - f(x)|$ for "most" $x$. That is, if $E_n = \{x : |f_n(x) - f(x)| > ||f_n - f||_{L^\infty}\}$, what can you say about $E_n$? Can you use the sets $E_n$ to come up with a single set $E$ that works for the proof?
For the backward direction, you not only need to show that $f \in L^\infty$, but that $f_n \to f$ in $L^\infty$ norm, i.e. $||f_n - f||_{L^\infty} \to 0$. This is not very hard if you just look at the definition of the $L^\infty$ norm.
Would this do? $E=\cup E_n$. for the backward direction; $||f_n - f||_{L^\infty}=\sup_{E^c} |f_n-f|\lt \varepsilon$. Thanks. – Colin Nov 19 '11 at 16:49