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!
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.