The Fourier Inversion Formula is excellently proved in the following site(Theorem 1): https://candes.su.domains/teaching/math262/Lectures/Lecture02.pdf
At the end of the proof, it is written as follows:
$$ I_{\epsilon}(t) = \int_{-\infty}^{\infty} f(u)g_{\epsilon}(t-u)du, $$ where $\int_{-\infty}^{\infty}g_{\epsilon}(x)dx = 1$ and $g_{\epsilon}(x)$ concentrates aound the origin, i.e., $x = 0$ as $\epsilon \mapsto 0$.
Q. Why $I_{\epsilon}(t) \mapsto f(t)$ as $\epsilon \mapsto 0$?
One reasoning is that $g_{\epsilon} \mapsto \delta$ as $\epsilon \mapsto 0$, where $\delta$ is the Dirac's delta function. However, I want to know an ``elementary'' but still rigorous proof of the above question.