# Fatou for weak convergence

I want to do exercise 3.2.4 from Rick Durett, Probability: Theory and Examples page 86.

$$\text{Let } g \geq 0 \text{ be continuous. If }X_n \Rightarrow X_{\infty} \text{ then } \liminf_{n\rightarrow \infty} \mathbb E(g(X_n))\geq \mathbb E(g(X_{\infty}))$$

My attempt:

Because $$X_n\Rightarrow X_{\infty}$$ there exists a random variabel $$Y_n$$ (with the same distribution as $$X_n$$) which converges to another random variable $$Y_{\infty}$$ almost surely.

So, we have $$\liminf_{n \rightarrow \infty}\mathbb E(g(X_n))=\liminf_{n \rightarrow \infty}\mathbb E(g(Y_n))\geq \mathbb E(g(Y_{\infty}))$$ by fatou and the continuity of g.

But can we say that $$\mathbb E(g(Y_{\infty}))=\mathbb E(g(X_{\infty})$$, if yes, then the prove would be finish.

It follows from the very definition that $X_n \to X_{\infty}$ in distribution is equivalent to $Y_n \to X_{\infty}$ in distribution for any sequence $(Y_n)_{n \in \mathbb{N}}$ such that $Y_n \sim X_n$.
Now in your case, as $Y_n \to Y_{\infty}$ almost surely, we have in particular $Y_n \to Y_{\infty}$ in distribution and therefore $X_{\infty} = Y_{\infty}$ in distribution.
• Hi. From your answer I guess that the condition $Y_{\infty} \sim X_{\infty}$ is needed, but this is not part of the assumption. (2nd line) Feb 1, 2015 at 18:16