TITLE: Dominated Convergence Theorem with "Almost Surely" replaced by "Convergence in Probability"
QUESTION [9 upvotes]: I want to show that if $\{X_n\}$ is a sequence of random variables such that:
(1) $\exists X$ (measurable) such that $X_n \xrightarrow{P} X$
(2) $\exists Y$ with $E(|Y|) < \infty$ such that $|X_n| \le Y \quad \forall n \in \mathbb{N}$
Then $E(X_n) \rightarrow E(X)$ and $E(|X_n - X|) \rightarrow 0$
How do I go about showing this?  This is of course the standard dominated covergence theorem changing almost sure convergence to $X$ to convergence in probability.  The approach to proving the standard theorem involves showing that $X$ is integrable and subsequently bounding $E(|X_n - X|)$, but I don't even understand how to do that part in this new set of assumptions.  Can someone help with this?  Once that part is done then the theorem is complete I believe?  I'm aware this might be a duplicate but I cannot understand these things very well and I'm very confused.  Thanks for any help you can give!

REPLY [5 votes]: Since $X_n$ converges in probability to $X$, it has a subsequence $X_{n_k}$ that converges almost surely to $X$. Letting $k\to \infty$ in $|X_{n_k}|\leq Y$ proves that $|X|\leq Y$, hence $X$ is integrable.
As a result, $|X_n-X|\leq 2Y$, hence $E(|X_n-X|)$ is bounded. Let us prove that the sequence $E(|X_n-X|)$ has $0$ as its only accumulation point. Suppose $E(|X_{m_k}-X|)$ converges to some $\ell$. Since $(X_{m_k})_k$ still converges in probability, it has a subsequence $(X_{m_{p_j}})_j$ that converges almost surely to $X$, hence $|X_{m_{p_j}}-X|$ converges a.s. to $0$ and $|X_{m_{p_j}}-X|\leq 2Y$. The dominated convergence theorem applies and yields $$\lim_j E(|X_{m_{p_j}}-X|) = 0$$
Hence $\ell=0$ and we conclude $\lim_n E(|X_n-X|) = 0$. By the reverse triangle inequality, $\lim_n E(X_n) = E(X)$.