# Does $X_n \to X$ in probability imply $\mathbb{E}(\liminf_n X_n) = \mathbb{E}(X)$?

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?

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