Let $\mu_n , $ be probability measures on $( \mathbb{R}, \mathcal{R})$ with $n \geq 1$ with charachterstic functions ${\Phi}_n$.
$\mu$ is also a probability measure with function $g$
Given that
$|{\Phi}_n (t)| \leq g(t)$ $\forall t \in \mathbb{R}$
and $\int_{-\infty}^\infty g(t)dt< \infty $
If $\mu_n \rightarrow \mu$ Show that ,
$\sup _{A\in \mathbb{R}}|\mu_n -\mu |\rightarrow0$ (i.e $\mu_n$ converges in $\mu$ in total variation norm)
My thought was to use try to use Levy continuity theorem. Or maybe sheffe's Theorem (See below).
But I am not sure how.
Best Answer
By inversion formula for characteristic functions the fact that the characteristic functions are integrable implies that they are absolutely continuous. Note that the characteristic function of $\mu$ is also dominted by the same integrable function $g$ so $\mu$ also has a density. Let $f_n$ and $f$ be the densities of $\mu_n$ and $\mu$. Then $f_n \to f$ a.e. by DCT applied to the inversion formula. Finally Scheffe's Lemma show that $\int |f_n-f| \to 0$. Of course, $|\mu_n(A)-\mu(A)| \leq \int |f_n-f|$ for all $A$.
Inversion formula: if characteristic function $\phi$ of $\mu$ is integrable then $\mu$ has density $f$ given by $f(x)=\frac 1 {2\pi} \int e^{-itx} \phi(x)dx$.