Why $\mathbb E[e^{itX}]$ determine uniquely the law of $X$? Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space and $X$ a random variable. Why $\mathbb E[e^{itX}]$ determine uniquely the law of $X$ ? And what does it mean ? (I'm not sure of the sense of "it determine uniquely the law of $X$).
 A: If $f(x)$ is the PDF of a continuous random variable $X$, $$\varphi(t):=\Bbb E\exp itX=\int_{\Bbb R}f(x)\exp itx dx.$$This is one definition of a Fourier transform of $f$. Fourier transforms are invertible, viz. $$f(x)=\frac{1}{2\pi}\int_{\Bbb R}\varphi(t)\exp-itxdt.$$There are some subtleties to a rigorous proof this works for more general distributions, such as expressing PMFs as Dirac combs, but the main point to why the characteristic function specifies a distribution is the above inversion theorem.
A: It mean that if $\mathbb E[e^{itX}]=\mathbb E[e^{itY}]$ for all $t$, then $\mathbb P\{X\leq x\}=\mathbb P\{Y\leq x\}$ for all $x$.
Let $f_X$ the density function of $X$. Then, $\mathbb E[e^{itX}]$ is the (inverse) Fourier transform of $X$. As you perhaps know, if $\hat f_X=\hat f_Y$ then $f_X=f_Y$ a.e. and thus, $$\mathbb P\{X\in A\}=\int_Af_X=\int_A f_Y=\mathbb P\{Y\in A\},$$
i.e. it uniquely define the law of $X$ (in the sense that if two random variable have the same characteristic function, they have the same distribution). 
It's a bit more subtly since a r.v. may be not continuous, but the idea is really this.
A: What it means is that for every Borel set $B\subseteq\mathbb R$ we can find the probability $P(X\in B)$ if  the function $\phi_X:\mathbb R\to\mathbb C$  prescribed by $t\mapsto\mathbb Ee^{itX}$ is at our disposal.
That fact can be stated as: the characteristic function $\phi_X$ determines the distribution of $X$.
On the other hand the distribution of $X$ determines the characteristic function $\phi_X$.
So actually there is a one-to-one relation between characteristic functions of random variables and distribution of random variable.
They both contain exactly the same amount of information about $X$.

You ask "why?" Well, it is just a fact and a more appropriate question would be: "how to prove that?".
For that it is actually enough to show that:


*

*the chararacteristics function of every random variable exists.

*the characteristic functions of $2$ random variable having distinct distributions are distinct.


I will not prove it here, but if you want a proof then have a look at wikipedia here.
