Probabilistic dominated convergence theorem

__Dominated convergence theorem:__ Suppose Xn->X almost surely and |Xn| ≤ W for all n, where E(W)<∞. Then E(Xn)->E(X).

1) Suppose Xn->X in probability and |Xn| ≤ W for all n, where E(W)<∞. Show that E(Xn)->E(X).

2) Suppose Xn->X in probability and (Xn)^2 ≤ W for all n, where E(W)<∞. Show that Xn->X in mean square. (i.e. E[(Xn-X)^2]->0)

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1) Let E(Xnk) be any subsequence of {E(Xn)}.

Then Xnk->X in probability => there exists a further subsequence Xnk' such that Xnk'->X almost surely.

By dominated convergence theorem, this implies that E(Xnk')->E(X) and so every subsequence of {E(Xn)} has a further subsequence of it which converges to E(X) => E(Xn)->E(X).

2) But now I am stuck with question 2. The assumption is replaced by (Xn)^2 ≤ W, and we need to prove more: Xn->X in mean square. I tried to modify the proof in question 1, but it doesn't seem to work here. How can we modify the proof? Or do we need something new?

Any help is much appreciated! :)