The closure of $C_c$ in $L^\infty$ is $C_0$ 
Let $X$ be a locally compact Hausdorff space that is $\sigma$-compact, and let $\mu$ be a Randon measure. Show that the closure of $C_c(X\to\mathbb{R})$ in $L^\infty(X,\mu)$ is $C_0(X\to\mathbb{R})$, the space of continuous real-valued functions which vanish at infinity. (i.e. for every $\varepsilon>0$ there exists a compact $K$ such that $|f(x)|\leq\varepsilon$ for all $x\in X\setminus K$).

Given $f\in C_0(X\to\mathbb{R})$, then for every $\varepsilon>0$  there is exist a compact set $K$ such that $|f(x)|\leq \varepsilon$ for all $x\in X\setminus K$. For fixed $\varepsilon>0$, we can define a function $g:\mathbb{R}\to \mathbb{R}$ by
$$g(x):=\begin{cases} x+\varepsilon,& \text{if}\ x< -\varepsilon,\\
0,&\text{if}\  -\varepsilon\leq x\leq\varepsilon,\\
x-\varepsilon,&\text{if}\ x>\varepsilon.
\end{cases}
$$
Then the function $h=g\circ f$ is continuous and supported on $K$ , thus $h\in C_c(X\to\mathbb{R})$. And we also have
$\|f-h\|_{L^\infty}\leq \varepsilon$. Now if we can prove $C_0(X\to\mathbb{R})$ is closed, we will complete the proof, but I have difficulty in proving this.
Let $f_n$ be a sequence in $C_0(X\to\mathbb{R})$ that converges to a limit $f$ in $L^\infty(X,\mu)$, then $f_n$ converges to $f$ uniformly outside of a null-set. How to show that there is a modification of $f$ on a null-set that is continuous and vanishs at infinity. The hint tell me to use Tietze extension theorem.
 A: In general $\left \| f \right \|_\infty \leq  \left \| f \right \|_u$ but if $f$ is continuous $\left \| f \right \|_\infty =\left \| f \right \|_u := \sup_{x \in X} |f|$, since $\lbrace x: |f(x)| > M \rbrace$ is clearly an open, and it's easy to verify that a not emtpy open has always positive measure, therefore $\mu (\lbrace x: |f(x)| > M \rbrace ) = 0$ if and only if $\lbrace x: |f(x)| > M \rbrace = \emptyset$, i.e. $\left \| f \right \|_\infty =  \left \| f \right \|_u$. Then it is sufficient to demonstrate the problem for $\left \| \cdot \right \|_u$.  We know that $C_c(X) \subset C_0(X)$. We show that the closure of $C_c(X)$ in the uniform norm or sup norm is $C_0(X)$. If  $\lbrace f_k \rbrace \subset C_c(X)$ is a sequence that converge uniformly to$f \in C(X)$ then for every $\epsilon >0$ exists $k \in \mathbb{N}$ such that $ \left \| f_k -f \right \|_u := \sup_{x \in X} |f_k -f| < \epsilon $, then $|f(x)| < \epsilon$ if $x \notin \mathrm{supp}(f_k)$, i.e. $f \in C_0(X)$. Conversely, if $f \in C_0(X)$, let the compacts $K_n=\lbrace x \in X : |f(x)| \leq 1/n \rbrace$ for $n \in \mathbb{N}$. Correspondingly to such compact there are functions $g_n \in C_c(X)$ with $0 \leq g_n \leq 1$ and $g_n=1$ on $K_n$. 
Then we define $f_n:=g_n f$, it's clear that $f_n \in C_c(X)$ and also $\left \| f_n -f \right \|_u < 1/n$, therefore $f_n \rightarrow f$ uniformly on $X$.
Note that we used substantially the fact that the $N_2$ (i.e. admits countable bases) - locally Hausdorff compact spaces - admit an exhaustive of compacts and in correspondence can be chosen of the Urysohn's function. This hypothesis has been replaced with the fact that the space $X$ is $\sigma-$compact, so it is not required that it is also $N_2$.
