# $\{X_1,X_2,...\}$ is uniformly integrable. Prove that $\mathbb{E}X_n \rightarrow \mathbb{E}X$ as $n \rightarrow \infty$.

Suppose $$X_n \rightarrow X$$ a.s. as $$n \rightarrow \infty$$ and that $$X_n \geq0$$ for all $$n, \omega$$. Furthermore, $$\{X_1,X_2,...\}$$ is uniformly integrable. Prove that $$\mathbb{E}X_n \rightarrow \mathbb{E}X$$ as $$n \rightarrow \infty$$.

I know that a sequence of random variables converges almost surely to $$X$$ if for every $$\epsilon > 0$$ we have $$$$P(\lim_{n \rightarrow \infty}|X_n - X|< \epsilon) = 1$$$$

But I'm not sure how to proceed. Some hints would be greatly appreciated.

• Both answers below are repeating the proof of Vitali's Theorem, which is worth looking into/knowing (and applies directly to your problem). Mar 2 '21 at 23:56
• True but, to be fair, this question is more-or-less asking to prove Vitali's theorem, so just stating this holds by Vitali's theorem wouldn't be very helpful. Although I agree, OP should read a proof of Vitali's theorem, as it is more general than what OP asked. Mar 3 '21 at 17:15

You can always write

$$\mathbb{E}\big[ \vert X_n-X \vert \big]= \int_F\vert X_n-X\vert dP+ \int_{F^c}\vert X_n-X\vert dP\leq \int_{F} \vert X_n\vert dP+ \int_F \vert X\vert dP+ \int_{F^c}\vert X_n-X\vert dP$$

Since $$X$$ is integrable and $$\{X_n\}$$ is unifromly integrable, for every $$\epsilon>0$$ there exists $$\delta_\epsilon>0$$ such that

$$\int_F\vert X_n\vert dP<\epsilon \quad \text{and} \quad \int_F\vert X\vert dP<\epsilon$$ if $$P(F)<\delta_\epsilon$$. Using almost sure convergence, we know that

$$P( \vert X_n-X\vert>\epsilon )\to0.$$

Therefore, for $$n$$ large enough we have $$P( \vert X_n-X\vert>\epsilon )<\delta_\epsilon$$ and therefore

$$\mathbb{E}\big[ \vert X_n-X \vert \big] \leq \int_{ \{ \vert X_n-X\vert>\epsilon \}} \vert X_n\vert dP+ \int_{ \{ \vert X_n-X\vert>\epsilon \} } \vert X\vert dP+ \int_{\{ \vert X_n-X\vert\leq\epsilon \}}\vert X_n-X\vert dP <$$

$$< 2\epsilon+ \int_{\{ \vert X_n-X\vert\leq\epsilon \}} \epsilon\cdot dP\leq 3\epsilon.$$

This is true for all $$\epsilon>0$$, so $$\mathbb{E}[X_n]\to \mathbb{E}[X]$$.

• What is the definition of the set $F$? Mar 7 '21 at 0:09
• @Math_Day $F$ is a general measurable set. In the last computation I take $F$ as $\{ \vert X_n-X\vert>\epsilon \}$. Mar 7 '21 at 6:46

Another (similar) approach is as follows.

For every $$\epsilon > 0$$, choose $$K(\epsilon)$$ so that $$E[|X_n|1_{X_n\geq K(\epsilon)}]<\epsilon$$ and $$E[|X|1_{X\geq K(\epsilon)}] < \epsilon$$ (this is simply applying the definition of UI to $$\{ X_n \}_n \cup \{X\}$$. Compute

\begin{align} E|X_n-X| \leq & E[|X_n-X|1_{X_n, X \leq K(\epsilon)}] \\ + &E[|X_n|1_{X_n > K(\epsilon), X \leq K(\epsilon)}] \\ +& E[|X_n|1_{X_n \leq K(\epsilon), X > K(\epsilon)}] \\ +& E[|X|1_{X_n > K(\epsilon), X \leq K(\epsilon)}]\\ +& E[|X|1_{X_n \leq K(\epsilon), X > K(\epsilon)}]\\ \leq &E[|X_n-X|1_{X_n, X \leq K(\epsilon)}] \\ + & E[|X_n|1_{X_n > K(\epsilon)}] \\ +&E[|X|1_{X > K(\epsilon)}] \\ +& E[|X_n|1_{X_n > K(\epsilon)}] \\ +&E[|X|1_{X > K(\epsilon)}] \\ \leq &E[|X_n-X|1_{X_n, X \leq K(\epsilon)}] + 4\epsilon. \end{align}

Sending $$n \to \infty$$ we find $$\limsup_n E|X_n-X| \leq 4\epsilon.$$ Now send $$\epsilon \to 0$$.