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 $$\lim_{n\rightarrow\infty}\int_{-\infty}^{\infty}(\hat{f_n}(s)-g(s))^2ds=0$$
$$\int_{-\infty}^{\infty}(g(s))^2ds=\int_{-\infty}^{\infty}(f(t))^2dt$$
How 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 $a \in R$
My result doesnt look pretty (looking forward to simpler one) but here is it: first prove it for $f$ with compact support then generalise it to $L^2$ using inequalities.
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 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