From the definition of total variation distance, we know that convergence in total variation implies weak convergence. However, suppose we have the following,
$$
d_{TV}(X_n, X) = Y_n,
$$
and $\mathbb{E}[Y_n] \rightarrow 0$, and hence, $\mathbb{E}[d_{TV}(X_n, X)] \rightarrow 0$.
If the expectation of the total variation distance converges to $0$, can we still somehow conclude that $X_n$ converges weakly to $X$?
$d_{TV}$ refers to the total variation metric. The total variation distance between two probability measures $P$ and $Q$ on a common probability space
$(\Omega, \mathcal{F})$ is given by,
$$
d_{TV}(P, Q) = \sup_{A \in \mathcal{F}} |P(A) – Q(A)|.
$$
Best Answer
First notice that $Y_n$ is a non-negative number, so $\operatorname{E}[Y_n]=Y_n$, then we have that $Y_n\to 0$. Now let $P_n$ the probability measure induced by $X_n$, and $P_X$ the probability measure induced by $X$ and set $Q:=\frac1{2}P_X+\sum_{n\geqslant 1}\frac1{2^{n+1}}P_n$, then its easy to check that $Q$ is a probability measure and that $P_n\ll Q$ and $P_X\ll Q$, therefore there are Radon-Nikodym derivatives $f_n$ and $f_X$ such that $P_n=f_n\cdot Q$ and $P_X=f_X\cdot Q$.
Let $\mathcal{B}(\mathbb{R})$ the Borel $\sigma $-algebra of $\mathbb{R}$ and note that
$$ \begin{align*} d_{TV}(X_n,X)&=\sup_{A\in \mathcal{B}(\mathbb{R})}\left|\int_{A}(f_n-f_X)dQ\right|\\ &=\int_{\{f_n-f_X\geqslant 0\}}(f_n-f_X)dQ\\ &=\frac1{2}\int_{\mathbb{R}}|f_n-f_X|dQ \end{align*} $$
where the last equality follows from the fact that
$$ \int_{\{f_n-f_X\geqslant 0\}}(f_n-f_X)dQ=\int_{\{f_n-f_X< 0\}}(f_X-f_n)dQ $$
as $\int_{\mathbb{R}}(f_n-f_X)dQ=0$. Therefore $f_n \xrightarrow{L_1}f_X$, what implies that $X_n\xrightarrow{\text{dist.}}X$ (as $L_1$ convergence implies weak convergence of measures).∎
An easier way to state the same is that for all $c\in \mathbb{R}$
$$ |P_n((-\infty ,c])-P_X((-\infty ,c])|\leqslant d_{TV}(X_n,X)\to 0 $$
so $X_n\xrightarrow{\text{dist.}}X$.