$X_n$ be a sequence of $L^1$ random variables converging a.s. to $X$ with $\inf_{n\ge 1} E(X_n) >-\infty$. To show $E(|X|)< \infty$

Question

Let $\{X_n\}_{n\ge1}$ be a sequence of random variables such that $X_{n+1}\le X_n,\forall n \ge 1$ and $ E(|X_n|) <\infty,\forall n \ge 1 $, $\inf_{n\ge 1} E(X_n) >-\infty$ and $X_n \to X$ almost surely.

Then how to show that $E(|X|)<\infty$ ?

Since $E(X_1)<\infty$, so applying MCT to $X_1-X_n$, we see that $E(X_n)\to E(X)$. Thus $|E(X)|<\infty$. But I can't quite see why $X \in L^1$. Please help.

Answer 1

Let $Y^+,Y^-$ denote the positive and negative parts of a variable $Y$, $Y=Y^+-Y^-$. A.s. convergence of the sequence implies the positive and negative parts converge a.s., $X_n^+\to X^+$, etc.

Since the sequence is decreasing, the positive part of the limit is integrable, $E(X^+)\le E(X_1^+)\le E|X_1|\le\infty$. The negative parts of the sequence, on the other hand, are increasing, so by monotone convergence $E(X^-)=\lim E(X^-_n)$, and we are finished if we show that $\lim E(X^-_n)<\infty$. That follows from

\begin{align} -\infty&< \inf_n E(X_n)=\inf_n(E(X_n^+)-E(X_n^-))\le \inf_n(E(X_1^+)-E(X_n^-))\\ &= E(X_1^+) - \sup E(X_n^-). \end{align}