Approximation for $L_{\text{loc}}^{\infty}(U)$ is this proof correct? Let $U\subset\mathbb{R}^n$ be open and bounded. I am trying to extend Evans' proof (in his PDE book) for approximating functions in $L_{\text{loc}}^{p}(U)$ for the case that $p=\infty$ using mollifiers. In particular I would like to show that
\begin{equation}%\label{eq: wts}
f\in L_{\text{loc}}^{\infty}(U)\Rightarrow f^{\varepsilon}\rightarrow f\text{ in } L_{\text{loc}}^{\infty}(U).
\end{equation}
I modified his proof as follows. Is it ok?
Let $V, W$ be open subsets of $U$ such that $V\subset\subset W\subset\subset U$. Now if $f\in L_{\text{loc}}^{\infty}(U)\Rightarrow f\in L_{\text{loc}}^{p}(U)$ for $1\leq p<\infty$. As in Evans (page 631) we have for sufficiently small $\varepsilon>0$
\begin{equation*}
\|f^{\varepsilon}\|_{L^{p}(V)}\leq\|f\|_{L^{p}(W)}.
\end{equation*}
We know that the Lebesgue measure is a Radon measure, $W$ is locally compact and Hausdorff and therefore $\overline{C_c(W)}=L^{p}(W)$ for $1\leq p<\infty$. So given any $\delta>0$ there is a $g\in C_c(W)$ such that:
\begin{equation*}
\|g-f\|_{L^{p}(W)}<\delta.
\end{equation*}
Then
\begin{align*}
\|f^{\varepsilon}-f\|_{L^{p}(V)}&\leq \|f^{\varepsilon}-g^{\varepsilon}\|_{L^{p}(V)}+\|g^{\varepsilon}-g\|_{L^{p}(V)}+\|g-f\|_{L^{p}(V)}\\
&\leq 2\|g-f\|_{L^{p}(W)}+\|g^{\varepsilon}-g\|_{L^{p}(V)}\\
&\leq 2\delta+\|g^{\varepsilon}-g\|_{L^{p}(V)}.
\end{align*}
Since $\mathcal{L}^{n}(V)<\infty$, 
\begin{equation*}
\|f^{\varepsilon}-f\|_{L^{\infty}(V)}\leq 2\delta+\|g^{\varepsilon}-g\|_{L^{\infty}(V)}
\end{equation*}
as $p\rightarrow \infty$.
Since $g^{\varepsilon}\rightarrow g$ uniformly on $V$, we have 
\begin{equation*}
\limsup_{\varepsilon\rightarrow 0}\|f^{\varepsilon}-f\|_{L^{\infty}(V)}\leq 2\delta.
\end{equation*}
This is true for all $\delta>0$ so we conclude 
\begin{equation*}
f^{\varepsilon}\rightarrow f\text{ in } L_{\text{loc}}^{\infty}(U).
\end{equation*}
 A: Each $f^\epsilon$ is continuous. If $f^\epsilon$ converges locally in $L^\infty$ then $f$ must be equal to a continuous function almost everywhere. There must be something wrong in your proof.
Consider this example: $U=(-1,1)\subset\mathbb{R}$, $f(x)=1$ if $0\le x\le1$, $f(x)=0$ if $-1\le x<0$ and $\phi(x)=(1-|x|)_+$. Then $\|f^\epsilon-f\|_{L^\infty(V)}=1/2$ for any open set $V\subset U$ containing $0$ (see the graph below.)

A: Let $U=(-2, 2),\ V=(-1, 1)$ and consider the signum function on  $U$ that is 
\begin{equation*}
\text{sgn}(x)=\left\{\begin{array}{l l}
1 &\quad x\in (0, 2)\\
0&\quad x=0\\
-1&\quad x\in (-2, 0)
\end{array}\right.
\end{equation*}
Then $V\subset\subset U$ and $\text{sgn}\in L^{\infty}(U)$ with $\|\text{sgn}\|_{L^{\infty}(U)}=1$. Furthermore, it is in $L_{\text{loc}}^{1}(U)$. It's mollifier, $\text{sgn}^{\varepsilon}$, is continuous in $V\subset U_{\varepsilon}$ for every $1>\varepsilon>0$. We claim that
\begin{equation}
\|\text{sgn}-\text{sgn}^{\varepsilon}\|_{L^{\infty}(V)}\geq\frac{1}{2}\quad \forall\ 1>\varepsilon>0.
\end{equation}
Suppose to the contrary. Then for some $1>\varepsilon>0$ we must have
\begin{equation*}
\|\text{sgn}-\text{sgn}^{\varepsilon}\|_{L^{\infty}(V)}<\frac{1}{2},
\end{equation*}
which means 
\begin{equation*}
-\frac{1}{2}+\text{sgn}^{\varepsilon}(x)< \text{sgn}(x)< \frac{1}{2}+\text{sgn}^{\varepsilon}(x)\quad\text{a.e. in }V.
\end{equation*}
Consequently, 
\begin{equation*}
\mathcal{L}\left(\left\{x\ \Big|\ \text{sgn}(x)\geq \frac{1}{2}+\text{sgn}^{\varepsilon}(x)\right\}\right)=0.
\end{equation*}
For this to be true we must have
\begin{equation*}
\frac{1}{2}+\text{sgn}^{\varepsilon}(x)\left\{\begin{array}{l l}
\geq 1&\quad x>0\\
\leq -1&\quad x<0
\end{array}\right.
\end{equation*}which clearly contradicts the fact that $\text{sgn}^{\varepsilon}(x)$ is continuous.
Note that we can replace the mollifier with any continuous function in $U$ and use the same argument to reach the same conclusion.
