A sequence of random variables with bounded variance 
If $\{X_n\}$ is a sequence of random variable with bounded variance:
  $$E|X_n^2|\le M<\infty,$$ and $X_n\to X$ in $L^1$, show that
  $$E|X^2|\le M.$$

I tried to use Fatou's lemma, Cauchy-Schwarz ... , but I failed to prove it. Can anyone give some hint?
 A: The following fact is well known (and strongly related to the Riesz-Fisher theorem):

If $X_n$ tends to $X$ in $L^1$, we can find a subsequence $X_{\phi(n)}$ such that $X_{\phi(n)}$ tends to $X$ both in $L^1$ and almost surely.

Applying Fatou's lemma to this subsequence yields
$$
E(|X|^2) = E(\liminf_{n\to\infty} |X_{\phi(n)}|^2) \leq \liminf_{n\to\infty} E(|X_{\phi(n)}|^2) \leq M.
$$
Therefore $E(|X|^2) \leq M$, and $E(|X|)^2 \leq M$ also holds because of Jensen's inequality (or Cauchy-Schwarz's if you prefer).
A: $$E|X|=E(|X-X_n+X_n|)\le E(|X-X_n|)+E|X_n|\ \forall n$$ Since $X_n\stackrel{1}{\to} X$, $\forall \epsilon>0$, for large enough $n$, $E|X-X_n|\le \epsilon$ Hence $$E|X|\le \epsilon +\sqrt{M}$$ for all $\epsilon>0\implies E|X|\le \sqrt{M} $.

In general, if $E|X_n^p|\le M$ and $X_n\stackrel{p}{\to}X$, $p\ge 1$, then $E|X^p|\le M$. This can be proved in a manner identical to the above, only using Minkowski's inequality.
A: $var \{X\}=E\{(X-E\{X\})^2\}$=$E\{X^2\}-(E\{X\})^2$.
  Taking the Schwarz's inequality $E^2\{|xy|\}<E\{|x|^2\}E\{|y|^2\}$ and setting 
y $\equiv 1$  we get $E^2\{|x|\}<E\{|x|^2\}$ so if $E\{|x|^2\}$ is bounded than $E^2\{|x|\}$ must be bounded
