1
$\begingroup$

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

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ Well, $g_n\to g$ in the $L^\infty$ norm. $\endgroup$ – user940 Dec 8 '15 at 13:48
  • 1
    $\begingroup$ $g$ is the limit of the uniformly-convergent continuous functions $g_n$, so $g$ is continuous. Take a look at the uniform limit theorem. $\endgroup$ – Pantelis Sopasakis Dec 8 '15 at 14:34
  • $\begingroup$ @PantelisSopasakis If a sequence of continuous functions $g_n$ converges to $g$, then converges to $g$ in $L^\infty$ norm? $\endgroup$ – Jeybe Dec 8 '15 at 15:22
  • $\begingroup$ What do we know about $E $? Is it open? Is it a subset of the closure of its interior? $\endgroup$ – PhoemueX Dec 8 '15 at 15:29
  • 1
    $\begingroup$ @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. $\endgroup$ – Pantelis Sopasakis Dec 8 '15 at 15:30
1
$\begingroup$

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.

EDIT: What I especially like about the above proof is that it works for any (Borel) measure on any topological space and not just for the Lebesgue measure. This is not true of many other arguments of which I tought about first.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.