IF $\mu_n \rightarrow \mu$ Show that , $\sup _{A\in \mathbb{R}}|\mu_n -\mu |\rightarrow0$

Question

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.

Answer 1

Here is a sketch of the proof of the statement along the lines of the OP's ideas:

• The assumptions on the characteristic functions imply that all the $\Phi_n$ are integrable.

• Since $\mu_n\stackrel{n\rightarrow\infty}{\Longrightarrow}\mu$, $\Phi_n\xrightarrow{n\rightarrow\infty} g$ pointwise (in fact uniformly in campsite sets ue to Lévy's continuity theorem).

• By dominate convergence ($|\Phi_n|\leq |g|$), $\|\Phi_n -g\|_{L_1}\xrightarrow{n\rightarrow\infty}0$.

• The Fourier inversion theorem, $\mu_n$ and $\mu$ are absolutely convergent with respect to Lebesgue's measure $\lambda$, and if $f_n=\frac{d\mu_n}{d\lambda}$ and $f=\frac{d\mu}{d\lambda}$ are the densities of $\mu_n$ and $\mu$, $$ \begin{align} f_n(y)&=\frac{1}{2\pi}\int e^{-ity}\Phi_n(t)\,dt\qquad \lambda\quad\text{a.s.}\\ f(y)&=\frac{1}{2\pi}\int e^{-ity}g(t)\,dt\qquad \lambda\quad\text{a.s.} \end{align} $$ Furthermore, the Fourier inversion theorem also implies the there are continuous functions $\bar{f}_n$ and $\bar{f}$ such that $f_n=\bar{f}_n$ and $f=\bar{f}$.

• Again, by dominated convergence $f_n\xrightarrow{n\rightarrow} f$ $\mu$-a.s.

• The conclusion now follows from Scheffe's theorem. (I provide a version of it and a short proof below)

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Theorem: Let $\{\mu,\mu_n\}_n$ be finite measures (positive) such that $\lim_n\mu_n(\Omega)=\mu(\Omega)$. Let $\nu$ be a $\sigma$--finite measure and suppose that $\mu_n,\,\mu\ll \nu$. If $\tfrac{d\mu_n}{d\nu}\rightarrow\tfrac{d\mu}{d\nu}$ $\nu$--a.s. then, $\|\mu_n-\mu\|_{TV}\rightarrow0$.

Proof:

If $\mu(\Omega)=0$, then the proof is trivial since $\|\mu_n\|_{TV}=\mu_n(\Omega)\rightarrow0$.

If $\mu(\Omega)>0$, let $\mu'_n= \tfrac{\mu(\Omega)}{\mu_n(\Omega)}\mu_n$ so that that $\mu'_n(\Omega)=\mu(\Omega)$. Hence, if $f_n=\tfrac{d\mu'_n}{d\nu}$ and
$f=\tfrac{d\mu}{d\nu}$, then $\int(f-f_n)_+\,d\nu=\int(f-f_n)_-\,d\nu=\tfrac12 \int|f-f_n|\,d\nu=\tfrac12\|\mu-\mu'_n\|_{TV}$.

Since $(f-f_n)_+\leq f$, dominated convergence implies that $\|\mu'_n-\mu\|_{TV}\rightarrow0$. As $$ \|\mu_n-\mu'_n\|_{TV}\leq \Big|\frac{\mu(\Omega)}{\mu_n(\Omega)}-1\Big|\sup_n\|\mu_n\|_{TV}\rightarrow0, $$ we conclude that $\|\mu_n-\mu\|_{TV}\rightarrow0$.

Answer 2

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 $f(x)=\frac 1 {2\pi} \int e^{-itx} \phi(x)dx$.