Characteristic function characterize the distribution Theorem: Let $\phi(t)=\int{e^{itX}dF_X}$ be a characteristic function of a random variable $X$. Then $$\displaystyle \lim_{T \to \infty}\int_{-T}^{T}{{\frac{e^{-ita}-e^{-itb}}{it}}\phi(t)dt}=P(X\in (a,b))+\frac{1}{2}(P(X=a)+P(X=b)).$$
I want to prove that if two random variables $X,Y$ have the same characteristic function, then they have the same distribution function. Obviously using the result. Obviously if $P(X=a)=0=P(Y=a) \: \forall a\in \Bbb R$ then the result is clear. 

Let $S$ denote the sets of points $a$ for which $\mathbb P\{X=a\}+\mathbb P\{Y=a\}\gt 0$. This set is at most countable. Hence, if $a\lt b$ are fixed, either $a,b\notin S$, and in this case $\mathbb P_X(a,b)=\mathbb P_Y(a,b)$, or at least one of these points is in $S$. If that happens, since $S$ is a countable set we can find $b_n\downarrow b$ such that $b_n\notin S$ and $a_n\uparrow a$ such that $a_n\notin S$ for all $n$. Then
$$\mathbb P(a \leq X \leq b)=\mathbb P\left(\bigcap_n \{X \in (a_n,b_n)\} \right)=\lim_n \mathbb P(a_n<X<b_n)=\lim_n \mathbb P_X(a_n,b_n),$$
and since $\mathbb P_X(a_n,b_n)=\mathbb P_Y(a_n,b_n)$ the desired result follows.
