$C^0$ is a closed subspace of $L^{\infty}$

Question

Let $\Omega\subset\mathbb{R}^n$ be an open bounded set. Let $f\in C^0(\bar\Omega)$. I have to prove that $\|f\|_{\infty}=\|f\|_{L^{\infty}}$. One implication is trivial. Let's consider the other one.

$$ \|f\|_{\infty}=\min\{K>0:|f(x)|\le K\,\,\,\, \forall x \in\bar\Omega\}\\ \|f\|_{L^{\infty}}=\inf\{K>0:|f(x)|\le K\,\,\,\,\,\mathrm{a.e. }x \in\Omega\}. $$

Let $K>0$ such that $|f(x)|\le K$ for every $x\in\Omega\setminus N$, where $|N|=0$. Since $f$ is continuous we have that $|f(x)|\le K$ for every $x\in\overline{\Omega\setminus N}$ (why?).

Then, I can conclude showing that $\bar\Omega\subseteq\overline{\Omega\setminus N}$.

Answer 1

Lemma. If $g\colon\mathbf R^n\to\mathbf R$ is continuous and $g(x)\leqslant 0$ for each $x\in A$, then $g(x)\leqslant 0$ for each $x\in \overline A$.

We use this with $g(x):=|f(x)|-K$ and $A:=\Omega\setminus N$.

To show the lemma, take $x$ in the closure of $A$ and $(x_n)_{n\geqslant 1}$ a sequence of elements of $A$ converging to $x$. Then $g(x_n)\leqslant 0$ for each $n$ and $\lim_{n\to +\infty}g(x_n)=g(x)$ by continuity of $g$.