# Show that : $\mathbb {E}(|X|)\leq \liminf_{n\rightarrow\infty}(\mathbb {E}(|X_n|))$

Let $$(\Omega,\mathcal{F},\mathbb{P})$$ be a probability space, $$\{X_n\}_{n\in \mathbb{N}}$$ is a sequence of random variables in $$L^1$$.which converges to $$X$$ in probability and a constant $$M>0$$ such that $$||X_n||_1\leq M$$.

Show that : $$\mathbb {E}(|X|)\leq \liminf_{n\rightarrow\infty}(\mathbb {E}(|X_n|))\leq \sup_{n}(\mathbb {E}(|X_n|))$$

I am thinking of applying the Fatou theorem to $$\{|X_n|\}_{n\in \mathbb{N}}$$, but I cannot think of a bound.

You don't require any bound on $$E|X_n|$$ for this. You can prove it using Fatou's Lemma.

Suppose $$E|X| >\lim \inf E|X_n|$$. Then we can find a real number $$c$$ and integers $$n_1 such that $$E|X| >c >E|X_{n_k}|$$ for all $$k$$. Since $$X_{n_k} \to X$$ in probability there is a subsequence, call it $$Y_j$$ which converges to $$X$$ almost everywhere. Bu then $$E|X| >c >E|Y_j|$$ for all $$j$$ and $$Y_j$$ converges to $$X$$ almost everywhere. This is a contradiction to Fatou's Lemma.

Choose a subsequence $$(n_k)_{j\geq 1}$$ such that

$$\liminf_{n\to\infty} \mathbb{E}[|X_n|] = \lim_{k\to\infty} \mathbb{E}[|X_{n_k}|]$$

holds. Since $$X_{n_k} \to X$$ in probability, we may find a further subsequence $$(n'_j)_{j\geq 1} \subseteq (n_k)_{k\geq 1}$$ such that $$X_{n'_j} \to X$$ almost surely as $$j\to\infty$$. Then by the Fatou's lemma,

$$\liminf_{n\to\infty} \mathbb{E}[|X_n|] = \lim_{j\to\infty} \mathbb{E}[|X_{n'_j}|] \geq \mathbb{E}\Bigl[ \liminf_{j\to\infty} |X_{n'_j}| \Bigr] = \mathbb{E}[|X|].$$

• Convergence in $L^{1}$ is not in the hypothesis. Commented Nov 5, 2019 at 23:58
• @KaboMurphy, Thank you for pointing out my mistake. I was careless when parsing OP's question. I fixed my answer accordingly. Commented Nov 5, 2019 at 23:59