First, I'm sorry is a bit too incoherent and too vague - I've only recently started learning about distributions and am just trying to get a general idea of what's happening.
There's a few things I would like to make sure I understand correctly:
1 We work with functions as distributions by identifying a function $f$ with the map $\psi \mapsto \int f \psi$. This is a well defined injection into the space of distributions.
2 We can naturally extend the definition of a fourier transform to tempered distributions by defining $F(T)(\psi) = T(F(\psi))$. To check that this is really an extension, we would like to show that $F(T_f) = T_{F(f)}$. The crucial observation here is that $\int f F(g) = \int F(f) g$ - this implies $F(T_f)(\psi) = \int f F(\psi) = \int F(f) \psi = T_{F(f)} (\psi)$.
3 Define $rect_n$ as $rect_n(x)=1$ for $x \in \langle -n,n \rangle$ and $0$ otherwise. Then the pointwise limit is $H=1$. Now apparently the fourier transform is continuous, so we should have that $F(H)(\psi) = \int H F(\psi) = \psi(0) = \delta (\psi)$ is the same thing as $\lim_{n \rightarrow \infty} F(rect_n)(\psi) = \lim_{n \rightarrow \infty} \int F(rect_n) \psi = \lim_{n \rightarrow \infty} \int F(rect_n) \psi$.
This should mean (I think) that $F(rect_n)$ approaches the dirac delta. But $F(rect_n) (\xi)= \int_{-n}^{n} e^{2\pi i\xi x} dx = \frac{\sin(2\pi \xi N)}{\pi \xi}$. But that doesn't appear to be the case - the function doesn't pointwise converge to $0$ for say $\xi = 1$.
Perhaps my conclusion that it converges to dirac delta is wrong, despite the fact that $\lim_{n \rightarrow \infty} \int F(rect_n) \psi = \psi (0)$. Or maybe I'm not working with the proper definition of continuity in this space.