# Show that the space of functions in $L^\infty (E)$ which admit a continuous representative is closed in $L^\infty (E)$

Let $E \subset \Bbb{R}^n$ be a set of positive measure. Let $\mathcal{C}$ be the set of measurable functions $f$ such that there exist a continuous $g$ with $f=g$ a.e. in $E \subset \mathbb{R}^n$. Prove that $\mathcal{C}$ is a proper closed subset of $L^\infty(E)$.

My attempt: if i set $\{f_n\}\subset \mathcal{C}$ a sequence of function with function limit $f$, then for each $n$ there exist $g_n$ continuous with $f_n = g_n$ a.e. but if $g_n \rightarrow g$ there's no guarantee that $g$ is continuous.

Then I tried with the limit points, but again I am stuck here. Any hint?

Thanks

• Well, $g_n\to g$ in the $L^\infty$ norm. – user940 Dec 8 '15 at 13:48
• $g$ is the limit of the uniformly-convergent continuous functions $g_n$, so $g$ is continuous. Take a look at the uniform limit theorem. – Pantelis Sopasakis Dec 8 '15 at 14:34
• @PantelisSopasakis If a sequence of continuous functions $g_n$ converges to $g$, then converges to $g$ in $L^\infty$ norm? – Jeybe Dec 8 '15 at 15:22
• What do we know about $E$? Is it open? Is it a subset of the closure of its interior? – PhoemueX Dec 8 '15 at 15:29
• @Jeremy For a continuous function $f$, its $L^\infty(X)$ norm is $\|f\|_\infty=\inf\{\beta \in [0,\infty]; \mu([|f|>\beta])=0\}$ $=\sup_{x\in X}|f(x)|$, thus, the uniform convergence of a sequence of continuous functions is convergence in the $L^\infty$ norm. – Pantelis Sopasakis Dec 8 '15 at 15:30

It is straightforward that $\mathcal{C} \subset L^\infty (E)$ is a subspace. We want to show that it is closed, which is equivalent to showing that it is complete. For this, it is equivalent to show that if a sequence $(f_n)_n$ satisfies $\sum_n \|f_n\|_\infty < \infty$, then $\sum_{n=1}^\infty f_n = \lim_{N\to\infty} \sum_{n=1}^N f_n$ exists in $\mathcal{C}$, also with convergence in $\mathcal{C}$, i.e. with respect to the $L^\infty$ norm. For a proof of this fact, see here: Completeness of the sum of two $L^p$ spaces.
Now, for each $n$, there is a continuous function $g_n$ with $f_n = g_n$ a.e. on $E$. Let us consider the modified ("truncated") function $$h_n := \max\{- \|f_n\|_\infty, \min\{ g_n, \|f_n\|_\infty\}\}.$$ Note that a.e. on $E$, we have $|g_n (x)| = |f_n(x)| \leq \|f_n\|_\infty$ and thus $h_n (x) = g_n (x)$. Furthermore, $h_n$ is continuous with $\|h_n\|_\sup \leq \|f_n\|_\infty$.
Now, it is well-known that the space $C_b (\Bbb{R}^n)$ of bounded continuous functions with the supremum norm (which is in general different from the $L^\infty$ norm!) is a Banach space. Since $h_n \in C_b (\Bbb{R}^d)$ and since $$\sum_n \|h_n\|_\infty \leq \sum_n \|f_n\|_\infty <\infty,$$ this implies that $h := \sum_n h_n$ is continuous and bounded, since the series converges uniformly.
It is now straightforward to see $h = \sum_n f_n$ a.e. on $E$, so that the $L^\infty$ function $f := \sum_n f_n$ is actually an element of $\mathcal{C}$.
Showing that $\mathcal{C}$ is a strict subset of $L^\infty$ is an easy exercise which I leave to you.