[From Thomas Ferguson, exercise $2.8$] Show if $X_n \rightarrow X$ a.s. and if $E(X_{n}^{2}) \rightarrow E(X^2) < \infty$, then $X_n \rightarrow X$ in quadratic mean.
So, I have to prove that $E\left(|X_{n} - X|^2 \right) \rightarrow 0.$ What I was thinking was something like
\begin{eqnarray} E\left(\left(X_{n} - X\right)^2\right) = E\left(\left(X_{n}^{2} - 2XX_n + X^2\right)\right) &=& E\left(\left(X_{n}^{2} - 2XX_n + 2X^2 - X^2\right)\right)\\ &=& E\left(\left(X_{n}^{2} - X^2 - 2XX_n + 2X^2\right)\right)\\ &=& E\left(\left(X_{n}^{2} - X^2\right)\right) - E\left(\left(2XX_n - 2X^2\right)\right). \end{eqnarray} Now I know that $E\left(\left(X_{n}^{2} - X^2\right)\right) \rightarrow 0$ from the hypotesis. But I don't know what to do with the other term and how to relate it with the a.s. convergence.
I know this problem may be related to the Scheffe or Riesz theorem (or dominated convergence), but I was thinking if this could be achieved using straightforward probability calculations without any result from measure theory. Thanks
Hint: Use Fatoo lemma with $$Y_n=2|X_n|^2+2|X^2|-|X-X_n|^2.$$