Let $(X_n)$ a sequence of random variables such that $X_n\xrightarrow[n\to+\infty]{law} 0$. I would like to show that $$\mathbb{E}\left(\log |1-X_n| \right)\xrightarrow[n\to +\infty]{} 0. $$ Intuitively, it seems natural. But since $x\mapsto \log |1-x|$ is not bounded on $\mathbb{R}$, it does not follow by definition of weak convergence.
Any hint/suggestion would be appreciated.
It doesn't follow. Say $(a_n)$ is a sequence of reals, and let $X_n$ be a random variable with $$p(X_n=0)=1-1/n,\quad P(X_n=a_n)=1/n.$$Then $X_n\to0$ in law regardless of the choice of $a_n$, but $$\mathbb{E}\left(\log |1-X_n| \right)=\log(|1-a_n|)/n$$need not tend to $0$.
For a maximally nasty example let $a_n=1$.