Fourier inversion formula for $L^2$ functions I have got a question about the Fourier-inversion Formula. Given a function $f \in L^2(\Bbb R)$ such that the following limit exists for  almost every $x\in \Bbb R$
\begin{equation}
\lim_{N \rightarrow \infty} \int_{-N}^N \hat{f}(y) e^{ixy} dy,
\end{equation}
where $\hat{f}$ is the Fourier transform. Define then
\begin{equation}
g(x):= \lim_{N \rightarrow \infty} \int_{-N}^N \hat{f}(y) e^{ixy} dy 
\end{equation}
How do $f$ and $g$ relate? I have read that Plancherel implies that $f=g$ almost everywhere, but i don't really know how to see that.
Thanks in advance!
 A: Let $g_N(x) := \int_{-N}^N \hat{f}(y) e^{ixy} dy$. Then according to your assumption $g(x)= \lim_{N \to \infty} g_N(x)$ pointwise a.e.
Note that $g_N(x) := \int_{-N}^N \hat{f}(y) e^{ixy} dy= \int_{\Bbb R} (\hat{f}.\Bbb 1_{[-N,N]})(y)e^{ixy}dy  $  . Now it is given that $f \in L^2(\Bbb R)$. Hence by Plancherel's theorem $\hat{f} \in L^2(\Bbb R)$. Which then implies that $\hat{f}.\Bbb 1_{[-N,N]} \in L^2[-N,N]$. Since, $[-N,N] \subset \Bbb R$ is compact , we get that $L^2[-N,N] \subset L^1[-N,N]$  thus $\hat{f}.\Bbb 1_{[-N,N]} \in L^1[-N,N]$. And from definition it follows readily that $\hat{f}.\Bbb 1_{[-N,N]} \in L^1(\Bbb R)$ , thus Fourier inversion is valid on $\hat{f}.\Bbb 1_{[-N,N]}, \forall N \in \Bbb N$.
Then it is clear that $g_N \to f$ in $L^2$ as $$\lim_{N \to \infty} ||f-g_N||_{L^2}^2 = \lim_{N \to \infty} \int_{|x| > N} |\hat{f}(y)|^2 dy =0$$ as we noted that $\hat{f} \in L^2(\Bbb R)$.
$g_N \to f$ in $L^2 \implies \exists$ a subsequence $\{g_{N_k}\}_{k \ge 1}$ such that $\lim_{k \to \infty}g_{N_k}(x)=f(x)$ a.e. Combining this with the our observation at the very beginning ( that $g(x)= \lim_{N \to \infty} g_N(x)$ pointwise a.e ) we obtain that $f=g$ a.e.
A: If you know that the Fourier transform is an isometric isomorphism in $L_2$ one may proceed as follows.
For any $k\in\mathbb{N}$ let $B(0;k)=\{x\in\mathbb{R}^d:\|x\|_2\leq k$ and $C(0;k)=[-k,k]^d$. For any $f\in L_2$, it is easy to check that $f_k=f\mathbb{1}_{C_k}, h_k=f\mathbb{1}_{B(0;k)}\in \mathcal{L}_1\cap\mathcal{L}_2$,  and
$$\lim_{k\rightarrow\infty}\|f-f_k\|_2=\lim_{k\rightarrow\infty}\|h_k-f\|_2\rightarrow0$$
Then, as $\widehat{f_k},\,\widehat{h_k}\in\mathcal{L}_2$ and
$$
\begin{align}
    \lim_{k\rightarrow\infty}\|\widehat{f}-\widehat{f_k}\|_2 =\lim_{k\rightarrow\infty}\|f-f_k\|_2=\lim_{k\rightarrow\infty}\|f-h_k\|_2=
    \lim_{k\rightarrow\infty}\|\widehat{f}-\widehat{h_k}\|_2 =0
    \end{align}
$$
That is,
$$
  \widehat{f}(y)=\lim_{k\rightarrow\infty} \int_{|x|_\infty\leq k}f(x)e^{-2\pi i x\cdot y}\,dx= \lim_{k\rightarrow\infty} \int_{|x|_2\leq k}f(x)e^{-2\pi i x\cdot y}\,dx
$$
in $\mathcal{L}_2$. Applying this to $widehat{f}$ leads to
$$
  \widehat{\widehat{f}}(y)=\lim_{k\rightarrow\infty} \int_{|x|_\infty\leq k}\widehat{f}(x)e^{-2\pi i x\cdot y}\,dx= \lim_{k\rightarrow\infty} \int_{|x|_2\leq k}\widehat{f}(x)e^{-2\pi i x\cdot y}\,dx
$$
The conclusion follows from the fact that  $\widehat{\widehat{f}}(x)=f(-x)$.
