This is an extension to the question I asked earlier here. The overarching question was the following homework problem,
Let $X_{n} \rightarrow X$ in probability. Show that $\liminf_{n} \mathbb{E}[X_{n}] \ge \mathbb{E}[X]$.
Obviously, this is very similar to the statement of Fatou's Lemma. I also know the following,
$X_{n} \rightarrow X$ in probability if and only if for all subsequences $X_{n(m)}$ of $X_{n}$ there exists a sub-subsequence $X_{n(m_{k})} \rightarrow X$ almost surely.
Given what was answered in my last question, it is trivial to show that Fatou's lemma applies to all of these sub-subsequences $X_{n(m_{k})}$, but how can I bring this back to $X_{n}$ itself, being that it lacks almost sure convergence to $X$?
EDIT Another condition is that $X_n \ge 0$
Best Answer
Edit: My first version was a bit too complicated, so here's a better version:
If $C = \liminf\limits_{n\to\infty}\;{\mathbb{E}[X_n]} = \infty$ there's nothing to prove, so assume that $C \lt \infty$.
We may choose a subsequence $X_{n(m)}$ such that $\mathbb{E}[X_{n(m)}] \to C = \liminf\limits_{n\to\infty}\;{\mathbb{E}[X_{n}]}$ by the definition of the $\liminf$. As you stated in your question, there's a sub-subsequence $X_{n(m_k)}$ such that $X_{n(m_k)} \to X$ a.e. since $X_n \to X$ in measure. As $X_n \geq 0$ a.e., the pointwise a.e. Fatou lemma gives us $\mathbb{E}[X] \leq \liminf\limits_{k \to \infty}\;E[X_{n(m_k)}]= \liminf\limits_{n\to\infty}\;{\mathbb{E}[X_n]}$, as desired.