Show that convergence in probabiltiy plus domination implies $L_p$ convergence

Question

I want to show that if random variable $X_n $ converges to $X$ in probability (Let $(\Omega, \mathcal{A},P)$ be the probability triple) and $|X_n| < Y \,\,\forall\, n$ then $X_n$ converges to $X$ in $L_p$.

Here's my attempt so far:

Since $|X_n-X| \leq |X_n|+|X|\leq Y + |X|$, I can use the dominated convergence theorem like so: $$\lim_{n \to \infty}\int_{\Omega} |X_n-X|^p dP = \int_{\Omega} \lim_{n \to \infty} |X_n-X|^p dP$$ This is the part where I want to use convergence in probability, but can't quite figure out how.

edit: Assumption: $Y \in L_p,$ as pointed out by the two answers below.

Answer 1

First of all, since $|X_n|\leq Y$ for all $n$, take a subsequence $(X_{n_k})$ of $(X_n)$ that converges to $X$ almost everywhere (this is possible due to convergence in probability) and taking limits in $|X_{n_k}|\leq Y$ conclude that $|X|\leq Y$ almost surely.

We will show that $\int|X_n-X|^p\to0$. Suppose that this isn't true, then there exists $\delta>0$ and a subsequence $(X_{n_k})\subset (X_n)$ such that for all $k$ it is $$\int|X_{n_k}-X|^p\geq\delta\;\;\;\;(*)$$

But $X_{n}\to X$ in probability, so $X_{n_k}\to X$ in probability too. Therefore we may find a subsequence $(X_{n_{k_j}})$ of $(X_{n_k})$ such that $X_{n_{k_j}}\to X$ almost everywhere. But it is also $|X_{n_{k_j}}-X|^p\leq(2Y)^p$, so by the dominated convergence theorem we must have that $$\int|X_{n_{k_j}}-X|^p\to0 $$ which is impossible due to $(*)$.

Edit: of course this is true under the assumption $Y\in L^p$. I assume you forgot to mention that.

Answer 2

Assuming $Y\in L_p$,

$$ \begin{align}E\big[|X_n-X|^p\big]&=E\big[|X_n-X|^p\mathbb{1}_{\{|X_n-X|>\varepsilon\}}\big]+E\big[|X_n-X|^p\mathbb{1}_{\{|X_n-X|\leq\varepsilon\}}\big]\\ &\leq2^pE\big[|Y|^p\mathbb{1}_{\{|X_n-X|>\varepsilon\}}\big] +\varepsilon^p \end{align}$$

Since $\lim_{n}P[|X_n-X|>\varepsilon]=0$ and $E[|Y|^p]<\infty$, the first integral in the second line converges to $0$ as $n\rightarrow\infty$ (This is due to the absolute continuity if the integral)