Does $X_n\overset{a.s.}{\to}X$ and $ \mathbf{E}(X_n^2) \to \alpha$ imply $\mathbf{E}(X^2) = \alpha$? I am wondering about the following: Does $X_n\overset{a.s.}{\to}X$ and $ \mathbf{E}(X_n^2) \to \alpha$ imply $\mathbf{E}(X^2) = \alpha$? I do not a priori know whether or not $X\in\mathcal{L}^2$. I think this is true and that the argument should be fairly simple but I cannot figure it out... 
I tried to do something like the following:
\begin{equation}
\mathbf{E}(X - \alpha)^2 \leq \mathbf{E} (X-X_n)^2 +\mathbf{E}(\alpha -X_n)^2 . 
\end{equation}
Now the last term becomes arbitrarily small when we let $n$ become large. Also,
\begin{equation}
    \mathbf{E} (X-X_n)^2 \leq  \mathbf{E}(X-X_n)^2 \mathbf{1}_{\{ |X-X_n|> \sqrt{\epsilon}  \}}+ \epsilon \mathbf{P}( |X-X_n|\leq \sqrt{\epsilon}).
\end{equation}
Can I reasonable continue from here?
Also, if $\mathbf{E}(X - \alpha)^2$ is indeed $0$, then apparently the conditions I have imply that $X$ is a.s. a constant. This seems like an overly strong result so I am suspicious.  
Thanks in advance for any clarification.
 A: Not true. On $(0,1)$ with Lebesgue measure let $X_n(t)=\sqrt n$ for $t\leq \frac 1 n$ and $0$ for $t >\frac 1 n$. Let $X=0$. Then $X_n \to X$ at every point, $EX_n^{2}=1$ for all $n$ but $EX^{2}=0$. 
A: To elaborate a little more, the random variables $X_n^2$ and $X^2$ satisfy $X_n^2 \overset{a.s}{\to} X^2$. We want the expectations to converge. This seems strictly weaker than convergence in $L^1$ of $X_n^2$ to $X^2$, but then by Scheffe's lemma we actually have their equivalence when the random variables are positive.
There is a condition , uniform integrability(read it up) defined for a sequence(or family) of random variables. Once it is done, (at least on $\mathbb R^d$, must hold in more generality), it turns out that almost sure convergence is equivalent to $L^1$ convergence for a u.i. sequence (to some random variable).
In the other answer, the given family is not uniformly integrable. I urge you to read the definition and check why this is the case from there. This also ends up implying the non-expectation of convergence, since the random variables are non-negative.
