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I could be making my life a bit too hard by raising this question, but the proof given by md2perpe on why $\lim_{\epsilon\downarrow 0}\frac{1}{\epsilon\sqrt{2\pi}}e^{\frac{-x^2}{2\epsilon^2}} = \delta(x)$, link: https://math.stackexchange.com/a/2834000/820472 is perfectly clear to me, except for the fact that in the end we obtain that $\lim_{\epsilon\downarrow 0}I_\epsilon = \psi(0)$. This is precisely $\left<\delta(0),\psi\right>$, however, we were tasked with showing that the said limit shows that the Dirac's delta given at any point $x$ is the weak limit of a certain Gaussian. Is it just implicitly assumed that one can reiterate the given proof with a modified Gaussian $\frac{1}{\epsilon\sqrt{2\pi}}e^{\frac{-(x - t)^2}{2\epsilon^2}}, t \in \mathbb{R}$, which then shows the equality at every $t \in \mathbb{R}$?

Epsilon Away
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    Yes. Of course. The last step you seek can simply be regarded as a change in the integration variable $x$. So it has little to do with the properties of the delta function. – M. Wind Sep 20 '22 at 17:16
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    Since $\mu_\varepsilon(dx)=\frac{1}{\varepsilon\sqrt{2\pi}}\exp(-x^2/(2\varepsilon),dx$ is a family of probability functions, one can also consider the characteristic functions $\widehat{\mu}_\varepsilon$ and see that they converge to $\widehat{\delta_0}(t)\equiv1$. – Mittens Sep 20 '22 at 18:01

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