Convergence in probability and $E(|X_n|) \to E(|X|)$ implies convergence in mean

Question

It is well known that in probability theory that, if $X_n$ converges in probability to $X$, the following are equivalent.

1. $X_n$ converges in mean to $X$
2. $E(|X_n|) \rightarrow E(|X|) < \infty$
3. $\{X_n\}$ is uniformly integrable

While I was able to find multiple proofs of the above results, all of them prove in the order of (1) -> (2) -> (3) -> (1).

I feel that it should be possible to prove (2) -> (1) directly without using the results for uniform integrability. For instance, there is a proof for $X_n \ge 0$ only using Dominated Convergence Theorem. Any pointers will be welcome.

Answer 1

The proof is surprisingly simple (if we have the right tools). It could be proved by using a variation of the Generalized Dominated convergence theorem, see this.

The theorem is stated without proof.

Theorem (Dominated convergence theorem).
Let $Z_n$, $Y_n$, $Z$ and $Y$ be R.V.'s such that $|Z_n| \le Y_n$ almost surely, $Z_n \overset{P}{\rightarrow} Z$, $Y_n \overset{P}{\rightarrow} Y$ and $E(Y_n) \rightarrow E(Y) < \infty$, then $E|Z_n-Z| \rightarrow 0$.

Question: Show that if $X_n \overset{P}{\rightarrow} X$ and $E(|X_n|) \rightarrow E(|X|) < \infty$, then $E|X_n-X| \rightarrow 0$.

Answer: Let $Z_n = |X_n - X|$ and $Y_n = |X_n| + |X|$ clearly that $Z_n \le |X_n| + |X| = Y_n$, $Z_n \overset{P}{\rightarrow} 0$, $Y_n \overset{P}{\rightarrow} 2|X|$ and $E(Y_n) = E|X_n| + E|X| \rightarrow 2 E|X| = E(2|X|) < \infty$. By applying the above DCT, we have $E|X_n-X| = E|Z_n - 0| \rightarrow 0$.