Show $\exists \{g_n\} \subset C_c^\infty$ such that $(f \ast g_n)(x) \to f(x)$ a.e., when $f \in L^1_\text{loc}$

Question

Let $C_c^\infty$ denotes the set of real valued function with compact support.
Show $\exists \{g_n\} \subset C_c^\infty$ such that $(f \ast g_n)(x) \to f(x)$ a.e., when $f \in L^1_\text{loc}$.

If $f$ were in $L^1$, the result would follow through Riesz theorem as $$ \|(f \ast g_n)(x)-f(x) \|_1 \to 0 \tag{1} $$ when $g_n=ng(nt)$ and for any $ g \in C_c^\infty$.

I don't think Riesz theorem applies in $L^1_\text{loc}$.
I tried to express $(f \ast g_n)(x)-f(x)$ in many ways, but none seems to lead me to the results.

Riesz Theorem: If $f_n$ goes to $f$ in $L^p(\mathbb R)$ then there exists a subsequence of $f_n$ converging pointwise a.e. to $f$.

This question is related.

Answer 1

You can apply the theorem to $f_m=f \phi_m$, where $\phi_m$ is a smooth function with compact support and $\phi_m \equiv 1$ on $[-m,m]$. Then $f_m \in L^1$, and you can find a sequence $(g_n)$ with $f_m * g_n \to f_m$ a.e. for $n\to\infty$. Assuming that the support of the whole sequence $(g_n)$ is contained in $[-1,1]$ (which is easy to achieve), you get that $f_m * g_n = f*g_n$ on $[-m+1,m-1]$, so $f*g_n \to f$ a.e. on $[-m+1,m-1]$. Since $m$ is arbitrary, the result follows.

The crucial point is that both the convolution itself and the result (a.e. convergence) are local, so $L^1_{\text{loc}}$ is as good as $L^1$ here.

Answer 2

If $f \in L_{\text{loc}}^1$, for any interval $[a,b]$, write $f$ as the sum of a function in $L^1$ and a function with support in $(-\infty, a-1] \cup [b+1,\infty)$.