A question on Fourier Inversion theorem

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I’m reading the proof of Fourier Inversion theorem:

$$\int_{\mathbb{R}^n}\widehat{f}(\xi)e^{2\pi ix\cdot\xi}d\xi=f(x)$$ almost everywhere, where $f,\widehat{f}\in L^1(\mathbb{R}^n)$.

The key is the equation $$\int_{\mathbb{R}^n}\widehat{f}(\xi)e^{2\pi ix\cdot\xi}e^{-\pi|\varepsilon \xi|^2}d\xi=\int_{\mathbb{R}^n}f(\xi)\varepsilon^{-n}e^{-\pi|\varepsilon^{-1}(\xi-x)|^2}d\xi,$$

where $\varepsilon$ is an arbitrary positive number. Let $\varepsilon\to 0^+$. I’ve understood the left side by a nice answer. Namely $$\lim_{\varepsilon\to 0^+}\int_{\mathbb{R}^n}\widehat{f}(\xi)e^{2\pi ix\cdot\xi}e^{-\pi|\varepsilon \xi|^2}d\xi =\int_{\mathbb{R}^n}\widehat{f}(\xi)e^{2\pi ix\cdot\xi}d\xi.$$ But I’m not sure if my reasoning is correct for the right side of the equation.

In the book it only said:

Let $\phi_\varepsilon(\xi)=\varepsilon^{-n}e^{-\pi|\varepsilon^{-1}\xi|^2}$. Then $\phi_\varepsilon$ is an approximate identity. We have $$\int_{\mathbb{R}^n}f(\xi)\varepsilon^{-n}e^{-\pi|\varepsilon^{-1}(\xi-x)|^2}d\xi=f*\phi_\varepsilon(x).$$

Yes, I know how to get the equation right above. My question is how to proceed. I know since $\phi_\varepsilon$ is an approximate identity and $f\in L^1(\mathbb{R}^n)$, we have

$$\Vert f*\phi_\varepsilon-f\Vert_1\to 0$$as $\varepsilon\to 0$.

It implies $$\int_{\mathbb{R}^n} |f*\phi_\varepsilon(x)-f(x)|dx\to 0$$ as $\varepsilon\to 0$. There is a theorem, although I forget what it’s called (by the way, what’s it called?), stating that

$E$ is a measurable set and $f$ is integrable on $E$. Then $$\int_E |f|~dm=0$$if and only if $f=0$ almost everywhere.

But I think that the integral converges to $0$ as $\varepsilon\to 0$ doesn’t mean it is $0$. It only means it’s infinitely small as $\varepsilon\to 0$. So it seems really close to the conclusion but I’m stuck here. How to refine my reasoning? Or could you give me your reasoning if mine is false?

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Since $f\star \phi_{\varepsilon}\to f$ in $L^1$, then there is a sequence $\varepsilon_k\to 0$ such that $f\star \phi_{\varepsilon_k}\to f$ almost everywhere in $\mathbb{R}^n$.