Convergence of sequence of Lebesgue integrals

62 Views Asked by At

Suppose we have a sequence of non-negative random variables $\{X_n\}_{n \in \mathbb{N}}$ which are integrable with respect to a probability measure $P$. Denote $\mu_n \equiv E(X_n)$ and suppose that $\mu_n$ converges to a limit in $\mathbb{R}$, say $\mu$.

Suppose $X_n \xrightarrow{d} X$. Is it true that $E(X) = \mu$?

It is true that $E(X)$ is well defined, as $|x|$ is a non-negative continuous function so that $E(|X|) \leq \liminf E(|X_n|) = \mu < \infty$ by the Portmanteau lemma.

I know that uniform integrability of the $X_n$ will guarantee the convergence, but I'm not sure if this is true or even needed for the problem. Any help would be massively appreciated!

1

There are 1 best solutions below

4
On

That is not true in general.

Take $f_n = n\cdot 1_{[0, 1/n]}$, which converges almost everywhere to $f = 0$. In particular, they converge in distribution as random variables on $[0, 1]$. However, we have

$$ \lim_{n\to\infty}\int f_n = 1 > 0 = \int f. $$

Aside:

After obtaining a Skorohod representation it boils down to Fatou's lemma. So it is hard to do better.