A proof that $EX_n\to EX$ for uniformly integrable $\{X_n\}$ with $X_n\to X$ a.s.

Question

I'm having some trouble following someone's proof of the following result:

Assume that $\{X_n\}$ are uniformly integrable and that $X_n\to X$ a.s.; then $EX_n\to EX$.

First, the author shows that $X$ is integrable, and I'm fine with that. But I don't understand how exactly she concludes that $$ \limsup_{n\to\infty}E|X_n-X|=0. $$

Fix $\epsilon>0$. By the integrability of $X$ and the uniform integrability of $\{X_n\}$, the author can choose an $M$ such that $$ \sup_n E\ (1_{\{|X_n|\ge M\}}|X_n|)\le\epsilon\quad\text{and}\quad E\ (1_{\{|X|\ge M\}}|X|)\le\epsilon. $$ I'm also fine with the identity \begin{align} |x-y| & \le 1_{\{|x-y|\le 2M\}}|x-y|+1_{\{|x-y|\ge 2M\}}|x-y|\\ & \le 1_{\{|x-y|\le 2M\}}|x-y|+(1_{\{|x|\ge M\}}+1_{\{|y|\ge M\}})(|x|+|y|). \tag{1} \end{align}

My question is, how exactly does this identity imply that \begin{align} \limsup_{n\to\infty}E|X_n-X| \le & \limsup_{n\to\infty}E(1_{\{|X_n-X|\le 2M\}}|X_n-X|)\\ &\ +2 \limsup_{n\to\infty}E(1_{\{|X_n|\ge M\}}|X_n|) + 2E(1_{\{|X|\ge M\}}|X|)? \tag{2} \end{align}

Specifically, how do the four terms in $$(1_{\{|x|\ge M\}}+1_{\{|y|\ge M\}})(|x|+|y|)$$ become $$2 \limsup_{n\to\infty}E(1_{\{|X_n|\ge M\}}|X_n|) + 2E(1_{\{|X|\ge M\}}|X|)?$$

EDIT: d.k.o. explains below why $$|x-y| \le 1_{\{|x-y|\le 2M\}}|x-y|+2|x|1_{\{|x|\ge M\}}+2|y|1_{\{|y|\ge M\}}, \tag{3}$$ from which $(2)$ is immediate. So, if the author claimed that $(2)$ follows from $(3)$, I'd have no question. Her claim in the proof, however, is that $(2)$ is a consequence of $(1)$. I'd still like to ask if this is actually true?

Answer 1

The inequality follows from the fact that $|x-y|\le |x|+|y|\le 2\{|x|\vee|y|\}\le 2\{|x|+|y|\}$. Namely,

$$|X_n-X|=|X_n-X|\cdot 1\{|X_n-X|\le 2M\}+|X_n-X|\cdot 1\{|X_n-X|> 2M\}$$ $$\le |X_n-X|\cdot 1\{|X_n-X|\le 2M\}+2[|X_n|\vee |X|]\cdot 1\{2[|X_n|\vee |X|]>2M\}$$ $$\le |X_n-X|\cdot 1\{|X_n-X|\le 2M\}+2|X_n|\cdot 1\{|X_n|>M\}+2|X|\cdot 1\{|X|>M\}$$

Answer 2

You can consider

$$ 1_{\{\vert x\vert>M\}} (\vert x\vert + \vert y\vert) = 1_{\vert x\vert >M} \vert x\vert + 1_{\vert x\vert >M}1_{\vert y\vert> M} \vert y\vert +1_{\vert x\vert >M}1_{\vert y\vert \leq M} \vert y\vert \leq 1_{\vert x\vert >M} \vert x\vert + 1_{\vert y\vert> M} \vert y\vert + 1_{\vert x\vert >M}1_{\vert y\vert \leq M} \vert y\vert $$

Similarly: $$ 1_{\{\vert y\vert>M\}} (\vert x\vert + \vert y\vert) \leq 1_{\vert x\vert >M} \vert x\vert + 1_{\vert y\vert> M} \vert y\vert + 1_{\vert y\vert >M}1_{\vert x\vert \leq M} \vert x\vert$$

So you can fix $\epsilon > 0$ an take $M$ such that $$\mathbb{E} (1_{\vert y\vert >M}1_{\vert x\vert \leq M} \vert x\vert) \leq \mathbb{E} (1_{\vert y\vert >M}\vert y\vert ) < \epsilon$$ and $$\mathbb{E} (1_{\vert x\vert >M}1_{\vert y\vert \leq M} \vert y\vert) \leq \mathbb{E} (1_{\vert x\vert >M}\vert x\vert ) < \epsilon$$

that should work (once $\epsilon$ is arbitrary)

Answer 3

Do you know Egoroff's theorem? Forgive me for rephrasing this result in non-probabilistic notation but here is how I'd state and prove it. We want to show that $f_n \to f$ in $L^1(X)$ for $\mu(X)=1$ and $f_n \to f$ almost everywhere.

Pick $\delta>0$. Egoroff's Theorem states that there is a set $A_{\delta} \subset X$ with $\mu(X\setminus A_{\delta}) < \delta$ when $\mu(X)<\infty$. This means we can write $$\int_X |f_n - f| d\mu \leq \int_{A_\delta} |f_n-f| d\mu + \int_{X \setminus A_\delta} |f_n| d\mu + \int_{X \setminus A_\delta} |f| d\mu$$ where the last two terms we got from the triangle inequality. Uniform integrability means these last two terms are less than $\epsilon>0$ each. So $$\int_X |f_n - f| d\mu \leq \int_{A_\delta} |f_n-f| d\mu + 2\epsilon.$$

By Egoroff's (remember), we can take the limsup of each side but pass the integral through the second integral term on the right side since the convergence is uniform. Hence

$$\limsup_{n \to \infty} \int_X |f_n - f| d\mu \leq 2 \epsilon.$$

But take $\epsilon \to 0$ to get the result.