Proof of Inversion Theorem for Characteristic Functions If $\mu$ is a probability measure and $\phi(t) = \int e^{itx}d\mu(x)$ is its characteristic function / Fourier transform, then we have that for $a<b$
$\lim_{T\rightarrow\infty}\frac{1}{2\pi}\int_{-T}^T \frac{e^{-ita} - e^{-itb}}{it}\phi(t)dt = \mu(a,b) + \frac{1}{2}\mu(\{a,b\})$
Could someone motivate the appearance of the function $1_{[-T,T]}\frac{e^{-ita} - e^{-itb}}{it}$ in this formula?  Is there a reason one might think to integrate $\phi$ next to this function in order to recover $\mu$?  Or is there some way to connect this formula to the inversion  formula for and $L^1$ function?
I am asking this because I believe I can verify each step of the proof of this result, but I feel that I may be missing the overall strategy.  
 A: $\mu$ is a finite positive measure. Note if $\mu$ is absolutely continuous then you can write $d\mu(x) = h(x) dx$ for some $h \in L^1$. 
But in general, $h$ is only a distribution so we need to regularize. Let $$f_n(x) = \int_{-\infty}^\infty  n e^{-\pi n^2 (x-y)^2}d\mu(y)= h \ast ne^{-\pi n^2 x^2}, \qquad \hat{f}_n(\omega)= \int_{-\infty}^\infty f_n(x)e^{-i\omega x}dx$$
$f_n$ is a Schwartz function so the Fourier inversion theorem applies 
$$f_n(x) = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{f}_n(\omega) e^{i\omega  x}d\omega$$
So that $$\lim_{\epsilon \to 0}\mu((a-\epsilon,b+\epsilon)) = \lim_{n \to 
\infty}\frac{1}{2\pi} \int_a^b f_n(x)dx = \lim_{n \to \infty}\frac{1}{2\pi} \int_a^b \int_{-\infty}^\infty \hat{f}_n(\omega) e^{i\omega x}d\omega$$
$$ = \lim_{n \to \infty} \frac{1}{2\pi} \int_{-\infty}^\infty \hat{f}_n(\omega) \int_a^b e^{i\omega x}d\omega=\lim_{n \to \infty}\int_{-\infty}^\infty \hat{f}_n(\omega) \frac{e^{i\omega b}-e^{i\omega a}}{2\pi i\omega }d\omega$$
And as $\phi(\omega) =\lim_{n \to \infty}\hat{f}_n(\omega) $, if it converges then it is $\displaystyle =\int_{-\infty}^\infty \phi(\omega) \frac{e^{i\omega b}-e^{i\omega a}}{2\pi i\omega }d\omega$
A: Suppose the measure has a smooth and fast decaying density. Then the measure of $[a,b]$ is the integral of the density on this interval. And the density is the inverse Fourier transform of the characteristic function.
Now the inverse transform is itself an integral, it is the integral of $\frac1{2\pi} e^{-itx}\varphi(t)$ with respect to $t$ on the whole real line. Switching the order of integration over $t$ and $x$ and solving the one over $x$ you get the magic term $\frac{e^{-itb}-e^{-ita}}{-it}$.
I hope this helps motivate.
Back to reality, without the assumption of smooth and fast decaying density, what you have is this limit in $T$ instead of an integral on the real line (it is important for the cancelations, this is called the principal value) and this average between $(a,b)$ and $[a,b]$.
A: The Fourier transform of $\chi_{[a,b]}$ is
$$
     \frac{1}{\sqrt{2\pi}}\int_{a}^{b}e^{-its}dt = \frac{1}{\sqrt{2\pi}}\frac{e^{-ibs}-e^{-ias}}{-is}.
$$
Therefore, by the Fourier inversion theorem, the inverse Fourier transform of the above function of $s$ converges to the mean of the left- and right-hand limits of the original characteristic function $\chi_{[a,b]}$. That is,
$$
    \lim_{T\rightarrow\infty}\frac{1}{2\pi}\int_{-T}^{T}\frac{e^{-ibs}-e^{-ias}}{-is}e^{isx}ds = \frac{1}{2}\left(\chi_{(a,b)}+\chi_{[a,b]}\right)
$$
When you apply this to the original expression, you get what you want.
