Expectation value of $Y \log Y$?

Question

I have a sequence $Y_n$ of real-valued random variables such that

1. $Y_n \ge 0$ almost surely, $\mathbb E[Y_n] =1$ and $\mathbb E[Y_n \log Y_n] \le 1$.

2. There exists a random variable $Y$ such that $\mathbb E[Y_n Z] \to \mathbb E[YZ]$, for any bounded random variable $Z$.

I need to prove that $\mathbb E[Y \log Y] \le 1$. How do I do that? I would like to show that $\mathbb E[Y \log Y] \le \liminf_{n \to +\infty} \mathbb E[Y_n \log Y_n]$, which I know it is true and it implies the conclusion. Problem is I can't see why this is true without obscenely overcomplicating the problem.

Answer 1

$\mathbb E[Y \log Y] \le \liminf_{n \to +\infty} \mathbb E[Y_n \log Y_n]$ follows since $x \log x$ is a convex function and therefore $\mathbb E[X \log X]$ is lower semicontinuous with respect to weak convergence in $L^1$.