TITLE: Convergence in Total Variation Implies Convergence in Distribution
QUESTION [5 upvotes]: Suppose $X,Y$ are random variables. We define the total variation distance of random variables to be
$d(X,Y)= \inf \{P(|X′−Y′|>0): X′,Y′$ are couplings of $ X,Y$ respectively$\}$.
Does convergence in total variation imply convergence in distribution? 
My feeling is that it should but I'm having trouble proving it.
Thanks!

REPLY [2 votes]: Let us use the dual representation of the total variation distance:
$$
d_{TV}(X,Y) = \sup_{f: \mathcal X \to [-1,1]} Ef(X) - Ef(Y)
$$
Then for every positive (measurable) bounded $f$:
\begin{align}
Ef(X_n)&\le Ef(X) +  \sup_{\mathcal X} |f|\times d_{TV}(X_n,X)\\
Ef(X_n)&\ge Ef(X) -  \sup_{\mathcal X} |f|\times d_{TV}(X_n,X)
\end{align}