Inverse Fourier Transform Formula on $L^2.$ Let $dt$ be the reduced Lebesgue measure on $\mathbb R,$ and $L^1$ and $L^2$ be the associated function spaces. There is an isometry $$f\mapsto \hat f:L^1\cap L^2\to L^2$$ given by $$\hat f(t)=\int f(x)e^{-ixt}dx.$$
Furthermore, since $L^1\cap L^2$ is dense in $L^2$, the map extends to an isometry from all of $L^2$ onto $L^2.$ 
The problem is to prove that if $f\in L^2$ and $\hat f\in L^1$ then the usual inversion formula holds a.e.: $$f(x)=\int \hat f(t)e^{ixt}dt.$$
We have, with $\chi_n$ the characteristic function on $[-n,n],\ \chi_n\cdot \hat f\to \hat f$ and since $\hat f\in L^1,$ the Dominated Convergence theorem applies to show that $$\lim \int \chi_n(t)\cdot \hat f(t)e^{ixt}dt=\int \hat f(t)e^{ixt}dt.$$ Thus, we have to show that $f_n(x):=\int \chi_n(t)\cdot \hat f(t)e^{ixt}dt\to f(x).$ 
The crux of this seems to be showing that $f_n(x)=\chi_n(x)\cdot f(x).$
Using the isometry, I have been able to show that $f_n\to f$ in the $L^2$ norm, so there is an a.e. pointwise convergent subsequence, but this seems like a dead end. 
If it turned out that $f_n$ was the inverse transform of $\chi_n\cdot \hat f,$ that would be enough also, but I do not see any reason for this to be true. The map $f\to \hat f$ maps $L^1\cap L^2$ onto a dense subspace of $L^2$, but there is no reason to suppose that it is $L^1\cap L^2$, or is there?
 A: Given an $L^2$ function $f(t)$, 
define $$\hat{f_n}(s)=\int_{-n}^{n}f(t)e^{-ist}dt$$  
It can be easily shown using Plancherel theorem that there is some $g(s) \in L^2$ and $$\lim_{n\rightarrow\infty}\int_{-\infty}^{\infty}(\hat{f_n}(s)-g(s))^2ds=0$$
$$\int_{-\infty}^{\infty}(g(s))^2ds=\int_{-\infty}^{\infty}(f(t))^2dt$$
We need to show $$\lim_{n\rightarrow\infty}\int_{-a}^{a}\left(\frac{1}{{2\pi}}\int_{-n}^{n}g(s)e^{ist}ds-f(t)\right)^2dt=0$$  where $a \in R$
If $ g (s) $ is in $ L^1$ then the convergence is pointwise ae
Proof for  $f(t) $ of compact support and $\in L^2$  :
clearly $f(t) \in L^1$ without loss of generality assume $f(t)=0$ outside $[-\pi,\pi]$ since we can always replace   $f(t)$ with $f(t')$ such as $t=\frac{n}{2\pi} t'$ for $n \in N$ and $t' \in R$.
define $${g}(s)=\lim_{n\to\infty}\int_{-n}^{n}f(t)e^{-ist}dt$$ 
The relationship between fourier transform and fourier series for $f(t)$ above follows from
$$(1)  \lim_{n\rightarrow\infty}\int_{-\pi}^{\pi}(\frac{1}{\pi}\int_{-\pi}^{\pi}f(t+u)\frac{sin(n+\frac{1}{2})u}{2sin(u/2)}du-\frac{1}{\pi}\int_{-\pi}^{\pi} f(t+u)\frac{sin(n+\frac{1}{2})u}{u}du)^2dt=0$$
which is simply consequence of dominated convergence theorem and Riemann-Lebesgue Lemma applied to the function $ f (t+u) k (u) $, where $k(u)=\frac{1}{2sin(u/2)}-\frac{1}{u}$  , using remainder term for the Taylor's series of $sin(u/2)$, it easily seen $k(u) $  is bounded over $[-\pi, \pi]$
$$(2)\frac{1}{2\pi} \int_{-(n+\frac{1}{2})}^{(n+\frac{1}{2})}g(s)e^{ist}ds=\frac{1}{\pi}\int_{-\infty}^{\infty}f(t+u)\frac{sin(n+\frac{1}{2})u}{u}du$$
This can be derived from fubini's theorem and  this theorem : $\mu$ is Lebesgue outer measure, $ A $  is a measurable set with finite measure. For all $n$ ,  $\int_Af_n^2d\mu \le k $ where $ k \in R $ , {$ f_n $} is uniformly integrable
Norm convergence for Fourier series  :
$$(3)  \lim_{n\rightarrow\infty}\int_{-\pi}^{\pi}(\frac{1}{\pi}\int_{-\pi}^{\pi}f(t+u)\frac{sin(n+\frac{1}{2})u}{2sin(u/2)}du-f(t))^2dt=0$$
using Riemann Lebesgue Lemma on (2) we have $$\frac{1}{2\pi} \int_{-(n+\frac{1}{2})}^{(n+\frac{1}{2})}g(s)e^{ist}ds=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t+u)\frac{sin(n+\frac{1}{2})u}{u}du$$
Using Norm convergence of fourier series (3), Riemann-Lebesgue lemma, Cauchy-Schwartz Inequality,$a^2+b^2 \ge 2ab$ and (1) and (2) we have:
$$\lim_{n\rightarrow\infty}\int_{-\pi}^{\pi}\left(\frac{1}{2\pi}\int_{-(n+\frac{1}{2})}^{n+\frac{1}{2}}g(s)e^{ist}ds-f(t)\right)^2dt=0$$
Proof for $f(t) \in L^2$:
define $f_m=f_{[-m,m]}$ where $m \in N$ and its fourier transform is $\hat{f_m}$
$$\lim_{m\rightarrow\infty} f_m=f$$ 
By Plancherel theorem : $\lim_{m\rightarrow\infty} \hat{f_m}=\lim_{m\rightarrow\infty}\int_{-m}^{m}f(t)e^{-ist}dt=\hat{f}$ in sense of $L^2$ 
our previous result for function with compact support : $$\lim_{n\to \infty}\int_{-\pi}^{\pi}(\frac{1}{2\pi}\int_{-n}^{n}\hat{f_m}(s)e^{ist}ds-f_m(t))^2dt =0 $$
$$\lim_{n\to \infty}\int_{-\pi}^{\pi}(\frac{1}{2\pi}\int_{-n}^{n}\hat{f_m}(s)e^{ist}ds-f_m(t))^2dt=\lim_{n\to \infty}\int_{-\pi}^{\pi}(\frac{1}{2\pi}\int_{-n}^{n}\hat{f_m}(s)e^{ist}ds-f_m(t)_{[-\pi,\pi]})^2dt =\lim_{n\to \infty}\int_{-\pi}^{\pi}(\frac{1}{2\pi}\int_{-n}^{n}\hat{f_m}(s)e^{ist}ds-f(t))^2dt=\lim_{n\to \infty}\int_{-\pi}^{\pi}(\frac{1}{2\pi}\int_{-n}^{n}\hat{f_n}(s)e^{ist}ds-f(t))^2dt=0 $$
There is a subsequence {$n$} such that $\hat{f_n} \to \hat{f} $ ae. Let's work with such subsequence 
$$\lim_{n\to \infty}\int_{-\pi}^{\pi}(\frac{1}{2\pi}\int_{-n}^{n}\hat{f}(s)e^{ist}ds-\frac{1}{2\pi}\int_{-n}^{n}\hat{f_n}(s)e^{ist}ds)^2dt =$$$$\lim_{n\to \infty}\int_{-\pi}^{\pi}(\frac{1}{\pi}\int_{-\infty}^{\infty}(f(t+u)_{[-\infty,-n]}+f(t+u)_{[n,\infty]})\frac{sin(n+\frac{1}{2})u}{u}du)^2dt $$ $$\le 2(||f(u)_{[-\infty,-n]}+f(u)_{[n,\infty]}||_{L^2}||\frac{sin(n+\frac{1}{2})u}{u}||_{L^2})^{-1/2}$$ by Cauchy-Scwartz ineqaulity
because $a^2+b^2 \ge 2ab$ , it follows from Cauchy-Schwartz inequality that:
$\lim_{n\to \infty}\int_{-\pi}^{\pi}(\frac{1}{2\pi}\int_{-n}^{n}\hat{f}(s)e^{ist}ds-f(t))^2dt\le$ 
$\lim_{n\to \infty}\int_{-\pi}^{\pi}(|\frac{1}{2\pi}\int_{-n}^{n}\hat{f}(s)e^{ist}ds-f_n(t)|+|f_n(t)-f(t)|)^2dt=0 $ 
we proved the result for some sequence {$n$} to see the result holds for all $n$ we observe that 
$$\lim_{n\to \infty}\int_{-\pi}^{\pi}(\frac{1}{2\pi}\int_{-(n+1)}^{(n+1)}\hat{f}(s)e^{ist}ds-\int_{-n}^{n}\hat{f}(s)e^{ist}ds)^2dt \le$$$$\lim_{n\to \infty}\int_{-\pi}^{\pi}(\frac{1}{2\pi}\int_{-(n+1)}^{(-n)}|\hat{f(s)}|ds+\frac{1}{2\pi}\int_{n}^{(n+1)}|\hat{f(s)}|ds)^2dt \le \lim_{n\to \infty} \frac{1}{2\pi}(||\hat{f}_{[-n-1,-n]}||_{L^2}+||\hat{f}_{[n,n+1]}||_{L^2})^2=0$$ by Hölder's Inequality
A: If $f\in L^2$, then $\hat{f}\in L^2$, and
$$
          L^2-\lim_n \frac{1}{\sqrt{2\pi}}\int_{-n}^{n}\hat{f}(s)e^{isx}ds = f.
$$
So there is a subsequence in $n$ so that the inverse integral over $[-n,n]$ converges pointwise a.e. to $f$. However, the pointwise inversion integral exists for all $x$ as $n\rightarrow\infty$ because $\hat{f}\in L^1$. Therefore,
$$
             f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat{f}(s)e^{isx}ds,\;\; a.e. x\in\mathbb{R}.
$$
If you knew $f$ were continuous everywhere, then equality would hold everywhere because a.e. equality for two continuous functions on $\mathbb{R}$ implies equality everywhere.
