Let $\Omega \subset \mathbb{R}^n$ be open set. Show that $C_0(\Omega)$ is not dense in $L^{\infty}(\Omega)$.
Notation :$f \in C(\Omega)$ and f has compact support then $f \in C_0(\Omega)$.
My attempt: Assume that $C_0(\Omega)$ is dense in $L^{\infty}(\Omega)$. Lets seek a contradiction. By the assumption for all $\varepsilon > 0$ there exist a function $g \in C_0(\Omega)$ such that $|f-g|_{\infty} < \varepsilon$. So we have $\mu(x \in \Omega: |f-g| > \varepsilon) = 0$. I tried to find an open subset of $\{x \in \Omega: |f-g| > \varepsilon\}$, but i failed. Any help would be apperciated.
Best Answer
Let $f$ be any discontinuous bounded measurable function on $\Omega$. (For example, the characteristic function of an open disk inside $U$). If there is sequence $(f_n) \subseteq C_0(\Omega)$ converging to $f$ in $L^{\infty}$ then $\|f_n-f_m\| \to 0$ which implies that $f_n$'s converge uniformly to $f$. But uniform limit of continuous functions is continuous.
[$L^{\infty}$ norm of a continuous function is same as its supremum norm].