# How to change the order of the limit and the expectation?

Does anyone know how to prove $$\lim E[X(n)]=E[\lim X(n)]$$???

Here I need to prove $$\lim E[X(n)]\le E[\lim X(n)]$$ and $$\lim E[X(n)]\ge E[\lim X(n)]$$.

Based on "Fatou Lemma", I can get that $$E[\liminf X(n)]\le \liminf E[X(n)]$$ and $$\limsup E[X(n)]\ge E[\limsup X(n)]$$.

Since we have $$\liminf E[X(n)] \le \lim E[X(n)]$$ and $$\lim E[X(n)] \le \limsup E[X(n)]$$, then we get $$E[\liminf X(n)]\le\lim E[Xn] \le E[\limsup X(n)].$$ Here I am stuck: if I can prove that $$E[\liminf X(n)]= E[\limsup X(n)]$$, it's done!! But I don't know how to prove it, does anyone know that??

$$\lim \mathbb{E}(X_n)$$ is not equal to $$\mathbb{E}(\lim X_n)$$ in general. Take
$$X_n=\begin{cases}n\hspace{1cm}\text{with probability }\dfrac{1}{n}\\0\hspace{1cm}\text{with probability }1-\dfrac{1}{n}.\end{cases}$$
Then $$X_n\stackrel{p}{\to}0,$$ but $$\mathbb{E}(X_n)=1$$ for all $$n.$$ That is, $$\lim \mathbb{E}(X_n)=1$$ but $$\mathbb{E}(\lim X_n)=0.$$