= Solution
Because the centered <normal distribution> density $g_t$ is nonnegative and has integral one, <Fubini's theorem> and translation invariance give
$$
\begin{aligned}
\|f*g_t\|_1
&\leq\int_{\mathbb R}\int_{\mathbb R}|f(x-y)|g_t(y)\,dy\,dx\\
&=\int_{\mathbb R}g_t(y)\|f\|_1\,dy
=\|f\|_1.
\end{aligned}
$$
Thus $f*g_t\in L^1$.
Use the angular-frequency convention
$$
\widehat f(\xi)=\int_{\mathbb R}f(x)e^{-ix\xi}\,dx.
$$
The <convolution theorem> and the Gaussian transform give
$$
\widehat{f*g_t}(\xi)=\widehat f(\xi)e^{-t\xi^2/2}.
$$
Since $|\widehat f(\xi)|\leq\|f\|_1$, this product is integrable. Moreover, absolute integrability justifies exchanging the integrals:
$$
\begin{aligned}
\frac1{2\pi}\int_{\mathbb R}\widehat f(\xi)e^{-t\xi^2/2}e^{ix\xi}\,d\xi
&=\int_{\mathbb R}f(z)
\left[\frac1{2\pi}\int_{\mathbb R}e^{-t\xi^2/2}e^{i(x-z)\xi}\,d\xi\right]dz\\
&=\int_{\mathbb R}f(z)g_t(x-z)\,dz\\
&=(f*g_t)(x).
\end{aligned}
$$
This proves the <Fourier inversion theorem> for the convolution directly.
Now additionally suppose $\widehat f\in L^1$. For every $x$, the <dominated convergence theorem> gives
$$
\lim_{t\downarrow0}(f*g_t)(x)
=\frac1{2\pi}\int_{\mathbb R}\widehat f(\xi)e^{ix\xi}\,d\xi.
$$
Fourier inversion identifies the right-hand side with $f(x)$ for <Lebesgue almost everywhere> $x$. Therefore the <Gaussian approximate identity> satisfies
$$
\boxed{f*g_t(x)\longrightarrow f(x)\quad\text{for almost every }x.}
$$
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