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My question is about the relation bwtween pointwise limit and convergence in probability.

It seems a basic question but I am stuck to it. Suppose $X_n \rightarrow X$ in probability

Then we konw $\liminf_{n \to\infty}X_n$ is not necessarily be equal to $X$

However, $$E[\liminf_{n \to \infty}X_n]= E[X]$$ Does the above equality holds? If so, how should I try to prove it?

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No, in general this fails to hold true. First of all, $X$ and $\liminf_{n \to \infty} X_n$ do not need to be integrable... but even if they are, we cannot expect this equality.

Consider the sequence of random variables $(X_n)_{n \in \mathbb{N}}$ on the probability space $((0,1],\mathcal{B}((0,1]))$ (endowed with Lebesgue measure $\lambda$) defined by $$\begin{align*} X_1(\omega) &:= -1_{\big(\frac{1}{2},1 \big]}(\omega) \\ X_2(\omega) &:= -1_{\big(0, \frac{1}{2}\big]}(\omega) \\ X_3(\omega) &:=- 1_{\big(\frac{3}{4},1 \big]}(\omega) \\ X_4(\omega) &:=- 1_{\big(\frac{1}{2},\frac{3}{4} \big]}(\omega)\\ &\vdots \end{align*}$$

It is not difficult to see that $\liminf_{n \to \infty} X_n = -1$ almost surely and $X_n \to 0$ in probability. Hence,

$$-1 = \mathbb{E}\left( \liminf_{n \to \infty} X_n \right) \neq \mathbb{E}(X)=0.$$

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