Q:

Pointwise evaluation of $L_2$ Fourier Transform

We know, that the $L_2$-Fourier Transform of a function $f\in L_2$ is defined as a limit of $L_2$ functions (e.g.  $\ \mathcal{F} f=\lim_{n\to \infty} \int_{-n}^{n} f\cdot \chi_{(-n,n)}\  d\lambda   $, convergence due to Plancherel)
We also know, that $L_2$ convergence does not imply pointwise convergence (there is only a subsequence converging pointwise a.e.)  $(\ast)$
My question now:
How is following equation justified under the conditions, that $f\in L_2(\mathbb{R})$, bandlimited (i.e. supp$(\mathcal{F} f) \subseteq  [-\lambda, \lambda]$ ): 
$$ (\mathcal{F}^{-1} \mathcal{F} f)(a)=\int_{-\infty}^{\infty} (\mathcal{F} f)(x)e^{2\pi i ax}dx=f(a) $$for arbitrary $a\in \mathbb{R}$.
The first equality is true, because $\mathcal{F}f \in L_1\cap L_2$ due to bounded support and therefore the (invers) $L_1$ and $L_2$ Transforms coincide a.e.
But although $\mathcal{F}^{-1} \mathcal{F} f=f$ a.e. and $f$ is countinous (Parley-Wiener), i am not sure, why we can evaluate the inverse Fourier transform pointwise? (as this would be in contradiction to $(\ast)$)
In order to illustrate my problem, i will present some "counterexample":
Take the functions $f_n=\chi_{A_n}\in L_2(0,1]$, where $A_n$ are smaller and smaller sets, that cover every point infinitely often, e.g. $(0,1], (0, 1/2], (1/2, 1], (0,1/4], (1/4, 2/4], (2/4, 3/4], (3/4, 1],
(0, 1/8], (1/8, 2/8]$ etc.
The $L_2$-limit function is the zero function, but the pointwise "evaluation", i.e. the pointwise limit $lim_{n\to \infty} \ f_n (a)$ for $a\in(0,1]$ doesn't even exist. 

A:

If I am reading your problem correctly, $f \in L^2$, and $\mathcal{F}f$ is band limited, which puts $\mathcal{F}f \in L^1\cap L^2$, and gives
$$
      \mathcal{F}^{-1}\mathcal{F}f = \frac{1}{\sqrt{2\pi}}\int_{-\lambda}^{\lambda}\mathcal{F}f(s)e^{isx}ds.
$$
And, it is also true that
$$
      f = \mbox{$L^2$-$\lim_{R\rightarrow\infty}$}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}\mathcal{F}f(s)e^{isx}dx.
$$
Therefore the $L^2$ functions $f$ and $\mathcal{F}^{-1}\mathcal{F}f$ are equal a.e.. If you know that $f$ is continuous, then
$$
    f(x) = \frac{1}{\sqrt{2\pi}}\int_{-\lambda}^{\lambda}\mathcal{F}f(s)e^{isx}ds.
$$
This is because both functions are continuous, and they're equal a.e..

