Convergence in probability implies convergence in $L^p$ with an additional condition

Question

I have a problem with this exercise.

Let $(X_n)$ be a succession of random variables such that $|X_n|<M \in \mathbb{R}$. Prove that if $(X_n)$ converges to a random variable $X$ in probability then $(X_n)$ converges to $X$ in $L^p$.

I tried to solve the exercise in the following way:

$X_n \to X$ in probability $\Rightarrow$ $X_n-X \to 0$ in probability $\Rightarrow$ $|X_n-X|^p \to 0$ in probability (because $|\cdot|^p$ with $p \ge 1$ is a continuous function).

I'm stuck here.

Thanks in advance for your help.

Answer 1

Hint: Use

$$\begin{align*} \int |X_n-X|^p \, d\mathbb{P} &= \int_{|X_n-X| \leq \epsilon} |X_n-X|^p \, d\mathbb{P} + \int_{|X_n-X| > \epsilon} |X_n-X|^p \, d\mathbb{P} \\ &\leq \epsilon^p + (2M)^p \mathbb{P}(|X_n-X|>\epsilon). \end{align*}$$

Remark: The statement is a particular case of Vitali's convergence theorem.