I know that if $F$ is a cdf then the inversion formula tells me that ($\phi$ is the cf of the rv corresponding to $F$) :
$F(b)-F(a)+ \displaystyle \frac{\mathbb{P}(X=b) - \mathbb{P}(X=a)}{2} = \lim_{T \rightarrow \infty} \displaystyle \int_{-T}^T \displaystyle \frac{e^{-ita}-e^{-itb}}{it}\phi(t)dt$
Suppose $X,Y$ are rv with same characteristic function , then I need to show that their cdf's are equal .
I have got that the cdf's must be equal at all common continuity points .
I know that I need to use the right continuity of cdf's here . How should I proceed ? Can it be shown that the set of common continuity points of $F $ and $G$ is dense if they are continuous ?
Let $F$ and $G$ be two distribution functions which coincide on a dense set $D$. Let $x \in \mathbb R$. There is a sequence $(x_n)$ contained in $D$ which decreases to $x$. Since $F(x_n)=G(x_n)$ for all $n$ and distribution functions are right continuous we get $f(x)=G(x)$.
To apply this property note that monotone functions have at most countable many points of discontinuity. Let $A$ and $B$ be the sets of discontinuity points of $F$ and $G$. Then $A\cup B$ is at most countable. Take $D$ to be $\mathbb R \setminus A\cup B$. This set is dense and $F$ and $G$ coincide at any point $x$ not in $D$.
The fact that $D$ is dense follows from the fact that any open interval is uncountable and hence it cannot be contained in $A\cup B$; thus it must contain at least one point of $D$.