# Proving $X_n\overset{\text{a.s.}}{\to}X \Rightarrow X_n\overset{L^1}{\to}X$ when $X_n$ are dominated by a r.v. with a finite mean

Let $X_n\overset{\text{a.s.}}{\to}X$, $|X_n|<Y$, and $E(Y)<\infty$. Then $X_n\overset{L^1}{\to}X$. Wikipedia says the statement without a proof. Could someone prove it or give a reference?

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Lebesgue dominated convergence theorem. – Did Jan 6 '13 at 16:07

## 2 Answers

When $X_n$ and $X$ are random variables defined on the measurable space $(\Omega,\mathcal F)$ and $X_n\to X$ almost surely, Lebesgue's dominated convergence theorem on the measured space $(\Omega,\mathcal F,\mathbb P)$ states that $\mathbb E(X_n)\to\mathbb E(X)$ (the limit of the expectations is the expectation of the limit) under the condition that $|X_n|\leqslant Y$ for every $n$ with $Y$ integrable. Hence the very same theorem also says that $\mathbb E(|X_n-X|)\to0$ under the same condition since then $|X_n-X|\leqslant2Y$ for every $n$.

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This is basically Lebesgue's dominated convergence theorem for random variables.

As answer to the comment: The expectation operator is usually defined to be an integral. So suppose we are working on the probability space $(\Omega,\Sigma,P)$, where $\Sigma$ is a $\sigma$-algebra and $P$ is a probability measure on $\Sigma$. Then the expectation of a random variable $X:\: \rightarrow \mathbb{R}$ is defined as $$\mathbb{E}X := \int_\Omega X\, \mathrm{d}P.$$ Now $X_n \to X$ a. s. and $(X_n)_n$ is dominated by an integrable random variable, so Lebesgue's theorem says we can interchange integral and limit, i. e. $\mathbb{E}X_n \to \mathbb{E}X$.

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Hmm...the theorem is about interchanging limit and integral sign. I can't imagine the proof. Could you give some more hints or details? – JDL Jan 6 '13 at 16:11
The expectation operator is usually defined to be an integral. So suppose we are working on the probability space $(\Omega, \Sigma, P)$, where $\Sigma$ is a $\sigma$-algebra and $P$ is a probability measure on $\Sigma$. Then the expectation of a random variable $X:\: \Omega \rightarrow \mathbb{R}$ is defined as $$\mathbb{E} X := \int_\Omega {X}\, \mathrm{d}P.$$ Now $X_n \to X$ a. s. and $X_n$ is dominated by an integrable random variable, so Lebesgue's theorem says we can interchange integral and limit, i. e. $\mathbb{E}X_n \to \mathbb{E} X$. – Thomas Jan 6 '13 at 16:19
@Thomas: your answer looks more like a comment, and your comment looks more like an answer. Maybe interchange them, or at least add your last comment to your answer? – Ilya Jan 8 '13 at 17:47
@Ilya You are right. I copied my comment to my answer. Thank you! – Thomas Jan 9 '13 at 8:19