$\mathcal{L}^1$ is closed under convergence in probability

Question

I'm dealing with the following statement:

Let $(X_n)$ be a sequence of random variables s.t $X_n \overset{p}{\rightarrow} X$ another random variable. Assume that $\sup_n E(\rvert X_n \lvert) < \infty$. Show that $X$ belongs to $\mathcal{L}^1$.

My sketch proof is to show that:

1. Since $X_n$ converges in probability to $X$, there exists a subsequence that converges almost surely to $X$

2. Since $\sup_n E(\lvert X_n \rvert)< \infty$, any element of the subsequence belongs to $\mathcal{L}^1$.

3. Finally, the subsequence of $X_n$ is a Cauchy sequence on $\mathcal{L}^1$ that converges to X. Since $\mathcal{L}^1$ is complete, we have that $X \in \mathcal{L}^1$

Can I follow this proof?

Answer 1

Your argument is not correct. How do you get a Cauchy sequence in $L^{1}$?. The result follows immediately from Fatou's Lemma: $E|X| \leq \lim \inf E|X_{n_{k}}| \leq \sup_n E|X_n|$ where $X_{n_{k}}$ is a subsequence converging almost surely.