If $|X_n| \le Y_n$ and $Y_n$ is integrable measurable R.V $\forall n \in \mathbb N$, and $Y_n$ converges to $Y$ in probability and $E[Y_n] \to E[Y] < \infty$, then by extending Fatou's Lemma, show that if $X_n$ converges to $X$ almost surley, this implies $X_n$ converges to $X$ in $L_1$.
I know that Fatou's Lemma states that if $Y \le X_n $ for $Y \in \mathcal L^1$ , then:
$$ E(\text{lim inf} X_n) = \text{lim inf} E(X_n)$$
Further, if $Xn$ converges to $X$ a.s. then
$$ P(\lim_{n\to\infty}(X_n=X))=1$$
and so we want to show $ E(|X_n-X|)$ converges to 0.
But what does it mean to "extend Fatou's Lemma"?! I'm not too sure how to answer this question…
Best Answer
First assume that $Y_n \to Y$ almost surely. Then $|X_n-X|\leq Y_n+Y$. Apply Fatou's Lemma to $Y_n+Y-|X_n-X|$. You get $2EY \leq \lim \inf (2EY-E|X_n-X|)$ (using the fact that $EY_n \to EY$). Hence $\lim \sup E|X_n-X| \to 0$.
For the general case go to subsequences. Use the fact $E|X_n-X|$ tends to $0$ iff every subsequence of this sequence has a further subsequence which tends to $0$ (and that convergence in probability implies almost sure convergence for a subsequence).