Exchanging limit and expectation for $L^2$ random variables Let $X_n$ be a sequence of random variables in $L^2$, i.e. $\mathbb E[\vert X_n \vert^2]<\infty$. Since the expectation value can be interpreted as a scalar product on $L^2$, can one exchange limit and expectation, without referring to theorems such as monotone convergence or dominated convergence, just using the continuity of the scalar product? :
$$\lim_{n\rightarrow \infty} \mathbb E[X_n] \stackrel{?}{=} \mathbb E[\lim_{n\rightarrow \infty} X_n]$$
 A: Continuity of scalar product implies $$\lim_{n \to \infty} \mathbb{E}((X_n -X) \cdot Y_n)=0$$ for all $(Y_n)_n \subseteq L^2$ such that $\sup_n \|Y_n\|_2 < \infty$ and $X_n \stackrel{L^2}{\to} X$. As @Did already wrote, the equality $$\lim_{n \to \infty} \mathbb{E}X_n = \mathbb{E} (\lim_{n \to \infty} X_n)$$ does not hold in general.
Consider for example the probability space $(\Omega,\mathcal{A},\mathbb{P}) := ([0,1],\mathcal{B}([0,1]),\lambda|_{[0,1]})$ and define $X_n$ by $$X_n(x) := n \cdot 1_{[0,\frac{1}{n}]}(x) \qquad (x \in [0,1])$$ Then you can easily show $\|X_n\|_2<\infty$, but $$1 = \lim_{n \to \infty} \underbrace{\mathbb{E}X_n}_{1} \not= \mathbb{E}(\underbrace{\lim_{n \to \infty} X_n}_{0})=0$$
A: Here is another way to show the exchangeability. Let $X$ be the $L^2$ limit of $\{X_n\}$. Then we have
$$\left| E[X_n]-E[X]\right| \leq E\left| X_n - X \right| \leq (E\left| X_n - X \right|^{\frac{1}{2}})^{\frac{1}{2}},$$
which implies that $E[X_n] \to E[X]$ as $n \to \infty$. The second inequality above is a special case of a corollary (see Corollary 3 in Real Analysis by Royden 4th Ed.(p142)):
"Let $E$ be a set of finite measure and $1\leq p_1 < p_2\leq \infty $. Then $||f||_{p_1}\leq c ||f||_{p_2}$ for any $f \in L^{p_2} (E)$, where $c=[m(E)]^{\frac{p_2-p_2}{p_2 p_1}}$ if $p_2<\infty$ and $c=[m(E)]^{\frac{1}{p_1}}$ if $p_2=\infty$."
Applying the collary to the current case where $m(E)=1$, $p_1=1$ and $p_2=2$, we can get the second inequality.
