Proof verification: Fourier Inversion theorem I want to prove Fourier Inversion theorem:
$$\int_{\mathbb{R}^n}\widehat{f}(\xi)e^{2\pi ix\cdot\xi}d\xi=f(x)$$
almost everywhere, where $f,\widehat{f}\in L^1(\mathbb{R}^n)$.
We can get a equation
$$\int_{\mathbb{R}^n}\widehat{f}(\xi)e^{2\pi ix\cdot\xi}e^{-\pi|\varepsilon x|^2}d\xi=\int_{\mathbb{R}^n}f(\xi)\varepsilon^{-n}e^{-\pi\varepsilon^{-2}|\xi-x|^2}d\xi$$
for any $\varepsilon>0$. For the left side of the equation, we apply the Lebesgue dominated convergence theorem. 

(Lebesgue dominated convergence theorem)$~~$Let $\{h_k\}$ be a sequence of measurable functions on a measurable set $E$. Suppose that the sequence converges pointwise to a function $h$ and is dominated by some integrable function $g$ in the sense that $$|h_k(x)|\le g(x)$$for all numbers $k\in\mathbb{N}_+$ and all points $x\in E$. Then $h$ is integrable and $$\int_E h(x)~dm=\lim_{k\to\infty}\int_E h_k(x)~dm.$$

In our case, let $$h(\xi):=\widehat{f}(\xi)e^{2\pi ix\cdot\xi}~~~\mbox{ and }~~~g(\xi):= |\widehat{f}(\xi)e^{2\pi ix\cdot\xi}|= |\widehat{f}(\xi) |$$
and we construct a sequence of measurable functions $\{h_k\}$ by $h_k(\xi):= \widehat{f}(\xi)e^{2\pi ix\cdot\xi}e^{-\pi|k^{-1}x|^2}$. Then clearly $$|h_k(\xi)|\le g(\xi)$$for all numbers $k\in\mathbb{N}_+$ and all points $\xi\in {\mathbb{R}^n}$. Since $g$ is also integrable, we have that 
$$\lim_{\varepsilon\to 0^+} \int_{\mathbb{R}^n}\widehat{f}(\xi)e^{2\pi ix\cdot\xi}e^{-\pi|\varepsilon x|^2}d\xi= \lim_{k\to \infty} \int_{\mathbb{R}^n} \widehat{f}(\xi)e^{2\pi ix\cdot\xi}e^{-\pi|k^{-1}x|^2}d\xi=  \left(\lim_{k\to \infty} e^{-\pi|k^{-1}x|^2}\right)\cdot\int_{\mathbb{R}^n} \widehat{f}(\xi)e^{2\pi ix\cdot\xi}d\xi= \int_{\mathbb{R}^n} \widehat{f}(\xi)e^{2\pi ix\cdot\xi}d\xi.$$
My question is: Is my reasoning right? I’m not sure about it. For example, the construction of $h_n(\xi)$ seems a little wired to me, but I think I must do it if I want to apply Lebesgue dominated convergence theorem. There was no sequence $\{h_n\}$ in our case originally, which is required in the dominated convergence theorem. Any help is appreciated, thanks!
 A: Your construction does not make very much sense to me. Here is how you can proceed:
If I understand your post you want to show
$$\lim_{\varepsilon\to 0^+} \int_{\mathbb{R}^n}\widehat{f}(\xi)e^{2\pi ix\cdot\xi}e^{-\pi|\varepsilon x|^2}d\xi= \int_{\mathbb{R}^n} \widehat{f}(\xi)e^{2\pi ix\cdot\xi}d\xi$$
Recall that $\lim_{x \to a} g(x) = L$ if and only if for every sequence $(x_n)_n$ in $\operatorname{dom}(g)\setminus \{a\}$ with $x_n \to a$, we have $g(x_n) \to L$.
We use this now.
So, let $0 < \epsilon_n \to 0$. We must show that
$$\lim_{n \to \infty} \int_{\mathbb{R}^n}\widehat{f}(\xi)e^{2\pi ix\cdot\xi}e^{-\pi|\varepsilon_n x|^2}d\xi= \int_{\mathbb{R}^n} \widehat{f}(\xi)e^{2\pi ix\cdot\xi}d\xi$$
For this, we can apply dominated convergence theorem. Indeed, first note that 
$$\lim_{n \to \infty} \widehat{f}(\xi)e^{2\pi ix\cdot\xi}e^{-\pi|\varepsilon_n x|^2}=  \widehat{f}(\xi)e^{2\pi ix\cdot\xi}
$$
for all $\xi$. 
Next, note that $$|\widehat{f}(\xi)e^{2\pi ix\cdot\xi}e^{-\pi|\varepsilon_n x|^2}| \leq |\hat{f}(\xi)|$$
for all $n$.
By your assumption, $\hat{f} \in L^1(\mathbb{R}^n)$ so we have found an integrable dominating function and the dominated convergence allows us to conclude.
