Fourier Transform Dirac Delta I have recently learnt about tempered distributions, and how one can define the Fourier transform of a tempered distribution $v$ to be $\hat v$ so that
$$\langle\hat v,\varphi\rangle=\langle v,\hat \varphi\rangle.$$
for all Schwartz function $\varphi$. Here, let's say I take the convention that the Fourier transform is
$$\hat\varphi(\xi)=\frac{1}{\sqrt{2\pi}}\int_\mathbb R e^{-i x\xi}\varphi(x) \, dx.$$
 In particular, one can show that the Fourier transform of $1$ is the Dirac delta $\delta_0$. Here, $1$ is in the sense of tempered distribution, defined by
$$\langle 1,\varphi\rangle=\int\varphi(x)\,dx.$$
No problem there.
However, I have come across a confusing usage of this fact: I was reading a proof of some estimate, the author claimed that ($g$ is just some function of $x$)
$$\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}\int_\mathbb{R} e^{-ixt}g(x)\,dt\,dx=\int_\mathbb{R}\delta(x)g(x)dx.$$
It looks like the author took the Fourier transform of $1$ "at the point $x$". But this is confusing to me because I don't understand what it means to take Fourier transform "at a point". I also don't understand what $\delta(x)$ means, and what it means to "integrate" it.
If the context is helpful, I am reading this set of seminar notes.
https://www.math.ucla.edu/~visan/Oberwolfach2012.pdf
Namely, page 11 Lemma 2.11 (Local Smoothing). I simplified the notation above the make the question less complicated.
 A: Actually it was (as you can check in the lecture notes you've referenced)
$$\frac 1 {2\pi} \int_{\mathbb R} \int_{\mathbb R} 
e^{-ixt} g(x)\ dt\ dx.$$
If you integrate by $t$ first then it can be rewritten as (I'll assume below that integration interval is always $\mathbb R$)
$$
 \int dx\ g(x) \left(\frac 1 {\sqrt{2\pi}}
\int  \frac{1}{\sqrt{2\pi}} e^{-ixt}\ dt
\right)
$$
Delta function is a generalized function that can be (roughly) defined as  such function that for any $f(x)$ $$\int dx\ f(x)\ \delta(x) = f(0),$$
so (inverse in your convention) Fourier transform of it is 
$$\hat{\delta}(t) = \frac{1}{\sqrt {2\pi}}\int dx\ e^{ixt\ }\delta(x) = \frac{1}{\sqrt{2\pi}}$$
and from this follows that 
$$\frac 1 {\sqrt{2\pi}}
\int  \frac{1}{\sqrt{2\pi}} e^{-ixt}\ dt =
\frac 1 {\sqrt{2\pi}}
\int  \hat{\delta}(t)\ e^{-ixt}\ dt
= \delta(x).$$
So $$\int dx\ g(x) \left(\frac 1 {\sqrt{2\pi}}
\int  \frac{1}{\sqrt{2\pi}} e^{-ixt}\ dt
\right) = \int dx\ g(x)\ \delta(x) = f(0).$$
A: Take g outside the inner integral on the left and you have the fourier transform of 1. This is the dirac delta, a distribution that integrates to 1 and is infinity at x. In this sense only the value of g where the dirac delta is non-zero becomes relevant. The point is then moving with x in your case.
A: Why so complicated?  The inner integral is  FT(g)(-t) and so we have to
look at the FT(FT(g))(x) for x = 0, but  FT(FT(g))(x) = g(-x) for test functions.
