Convergence of Random Variables in L1

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Let $\{X_n\}$ be a sequence of integrable, real random variables on a probability space $(\Omega, \mathscr{F}, \mathbb{P})$ that converges in probability to an integrable random variable $X$ on $\Omega$. Suppose $\mathbb{E}(\sqrt{1 + X_n^2}) \to \mathbb{E}(\sqrt{1 + X^2})$ as $n\to \infty$. Show that $X_n\xrightarrow{L^1} X$.

 

[Office_Shredder]

I have no solution attempt, but thought I would write some random stuff to get the conversation going:
Converges in probability means $P(|X_n-X|>\epsilon)\to 0$ for all $\epsilon>0$.

Converges in $L_1$ means $E(|X_n-X|)\to 0$. One example where these aren't the same is: $X$ is identically zero, $X_n$ is $n$ with probability $1/n$ and 0 otherwise. $P(|X_n-X|>\epsilon)\leq 1/n\to 0$, but $E(|X_n-X|)=1$ for all $n$.

The expected value condition is interesting, I wonder if the $1+$ piece is necessary.

 

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Here is a hint: Convergence in probability implies convergence almost surely.

 

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Spoiler

Let $\{X_{n_k}\}$ be a subsequence of $\{X_n\}$. Since $X_n\to X$ in probability, there is a further subsequence $\{X_{n_{k_j}}\}$ of $\{X_{n_k}\}$ that converges to $X$ almost surely. Now $|X_{n_{k_j}}| \le \sqrt{1 + X_{n_{k_j}}^2}$ and $\mathbb{E}(\sqrt{1+X_{n_{k_j}}^2}) \to \mathbb{E}(\sqrt{1+X^2}) < \infty$, so by the generalized dominated convergence theorem $X_{n_{k_j}} \xrightarrow{L^1} X$. Since $\{X_{n_k}\}$ is an arbitrary subsequence of $\{X_n\}$, the result follows.