Durett Probability Theory and Examples suggest that the following inversion result (p.95) is intuitive.
However, I cannot figure out how to prove it.
Here is the result :
If $X$ has characteristic function $\phi$, and distribution $\mu$, then : $$\mu(\{a\}) = \lim_{T \to \infty} \frac{1}{2T} \int_{-T}^{T}e^{-ita}\phi(t)dt$$
Any help appreciated thanks!
Similar to the proof of inversion formula, let $$I_T = \frac{1}{2T}\int_{-T}^T e^{-ita}\varphi(t)dt = \frac{1}{2T}\int_{-T}^T \int e^{-ita}e^{itx}\mu(dx)dt$$
By Fubini's theorem (as $e^{-ita}\ge 0$) and odd property of $\sin$ function, we have: $$I_T = \frac{1}{2T}\int\int_{-T}^T e^{-ita}e^{itx}dt\mu(dx) = \frac{1}{2T}\int\int_{-T}^T \cos(t(x-a))dt\mu(dx)$$ And we can easily observe that the integral $\int_{-T}^T\cos(t(x-a))dt$ is bounded when $x-a$ not equal to zero (because of its periodicity) when $T\to\infty$, and when $x = a$, the integral equal to $2T$. So we have: $$\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T\cos(t(x-a))dt = \left\{\begin{aligned} &0 , &x\not= a \\ &1 , &x = a \end{aligned}\right.$$ Then by BCT, we have $I_T\to\mu(\{a\})$.