Let $S$ be an open and bounded subset of $\mathbb{R}^n$ and let $f, g \in L^1_{\mathrm{loc}}\left(S\right)$ be arbitrary and $c \geq 1$. I am trying to show that there exists a sequence $(\varphi_k)_{k=1}^\infty\subset C_0^\infty(S)$ such that for all $x\in S$ we have $|\varphi_k(x)| \leq c$ and for Lebesgue almost every $x\in S$ we have $\lim_{k\to\infty}\varphi_k(x) = \mathrm{sgn(f(x) - g(x))}$ where
$$\text{sgn}(x) = \begin{cases}1 &: x > 0 \\ 0 &: x = 0 \\ -1 &: x < 0\end{cases}$$
For future reference, take $h(x) := \mathrm{sgn(f(x) - g(x))}$.
Since $C_0^\infty(S)$ is dense in $L^p(S)$ for all $p\geq 1$ and any $L^p$ convergent sequence has a subsequence that converges almost everywhere pointwise (to the $L^p$ limit), we may directly jump to such a sequence $(\varphi_k)_{k=1}^\infty\subset C_0^\infty(S)$ for which $\lim_{k\to\infty}\varphi(x) = h(x)$.
Question: What is unclear to me is how can I rescale such a sequence $(\varphi_k)_{k=1}^\infty$ to satisfy the condition $|\varphi_k|\leq c$ in $S$? My knee-jerk reaction would be to discard zero functions from the sequence $(\varphi_k)_{k=1}^\infty$ and to normalize each element $\varphi_k$ by its $L^\infty$ norm. However, such elements $\widetilde{\varphi_k}(x) := \frac{\varphi_k}{||\varphi_k||_{L^\infty(S)}}$ might fail to converge to $h$ when $h(x) = \pm 1$. Could this problem be solved by a suitable mollified?
No need even for mollifiers. Just let $\psi\in C^\infty_b(\mathbb R)$ be any smooth bounded function that fixes $0$, $1$, and $-1$, e.g. $\psi(x)=\sin(\frac{\pi}{2}x)$, then consider $\psi\circ \varphi_i$.
Remark
The proof that $C^\infty_0$ functions are dense in $L^p$ in the first place also shows that $C^\infty_0$ functions bounded by some $c\geq 0$ are dense in the subset of $L^p$ with the same essential bound. That is, an $L^p$ function essentially bounded by $c$ can be approximated by step functions bounded by $c$, and these can in turn be approximated by step functions with compact sets, and these finally can be approximated by linear combinations of bump functions with disjoint compact supports, which therefore maintain the same bound of $c$.