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I am reading the proof of the dominated convergence theorem

Dominated Convergence Theorem

Let $X$ and $X_{n}$, $n=1,2,3...$ be random variables. Suppose that there's a random variable $Y$ such that

(1)$|X_{n}|\leq Y$ a.s for all $n$

(2)$\lim_{ x\to \infty}$ $X_{n}=X$ either a.s or in probability

(3)$E(Y)< \infty$ Then$$\lim_{ x\to \infty}E(X_{n})=E(X)$$

I have a question about one sentence of the proof

Suppose $X_{n} \to X$ in probability. Since $E(|X_{n}|) \leq E(Y)$, the sequence of expectations is bounded. If it does not converge to $E(X)$, there exists a subsequence $X_{n_{k}}$ such that $E(X_{n_{k}})$ converges to a limit $L \neq E(X)$

My question is that, when we know the expecations are bounded, we only know the sequence has a convergent subsequence, how do we know that the limit of the subsequence is not $E(X)$?

Thanks in advance!

Community
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Brown
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    We don't. But we also don't know that the limit exists an is equal to $E(X)$. So there are two cases to consider: In one case, the theorem holds. In the other, we have to say something - likely proving a contradiction. – WoolierThanThou Jan 07 '20 at 13:25
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    If we assume that it does not converge, then there must be at least two different convergent subsequences with different limits, the details are given by this answer https://math.stackexchange.com/a/298828/631468 – Leander Tilsted Kristensen Jan 07 '20 at 13:41

2 Answers2

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If the sequence $E(X_{n})$ does not converge, then $\liminf_{n} E(X_{n}) < \limsup_{n} E(X_{n})$. So $\liminf_{n} E(X_{n}) \neq E(X)$ or $\limsup_{n} E(X_{n}) \neq E(X)$. For example, suppose $\liminf_{n} E(X_{n}) \neq E(X)$. Then there is a convergent subsequence converging to $\liminf_{n} E(X_{n})$.

Skipping the details, you can just take a convergent subsequence converging to $L \neq E(X)$.

starboy
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To prove it, you should recall a result:

If $X_n \to X$ in probability, then for every subsequence $n_k$ there exists a further subsequence $(n_k)_j$ that $X_{(n_k)_j} \to X$ almost surely.

In this proof I assumed that we know the version of DCT where we have the almost sure convergence assumption.

First of all as you mentioned $\mathbf{E}[|X_n-X|]$ is bounded. Consider $\limsup\mathbf{E}[|X_n-X|]$ and assume it is some number $L \geq 0$. Therefore, it means there exists a subsequence $n_k$ such that $\mathbf{E}[|X_{n_k}-X|]$ converges to $L$. Note that by assumption $|X_{n_k}-X|$ converges in probability to $0$. Therefore, there exists a further subsequence $(n_k)_j$ such that $|X_{(n_k)_j}-X|$ converges to zero almost surely and it is bounded by $2Y$ for all $n$. The "usual" DCT tells us that $\lim_{n\to \infty} \mathbf{E}[|X_{(n_k)_j}-X|]=0$. It is a contradiction with the fact that $L>0$. Therefore, $L$ has to be zero. As a result we have $$ \lim_{n \to \infty}\mathbf{E}[|X_n-X|]=0 $$ Therefore using the fact that $|X_n|\leq |X_n-X| + |X|$, we have $\lim_{n \to \infty}\mathbf{E}[|X_n|]=\mathbf{E}[|X|]$.

MMH
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