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I guess that convergence in distribution does not mean that the moments must be bounded. I need to find a toy example that confirm the above idea.

Mathematically, I want to find a sequence $\{X_n\}$ that each $X_n \xrightarrow{d} N(0,1) $ as $n \rightarrow \infty$ but $E\{ |X| \} = \infty$. (Here, $N(,)$ is normal distribution and $\xrightarrow{d}$ is convergence in distribution).

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    Consider $X$ standard normal, $Y$ any other random variable and $(B_n)$ a sequence of Bernoulli random variables independent of $(X,Y)$ with $P(B_n=1)\to1$, $P(B_n=1)\ne1$, then $X_n=B_nX+(1-B_n)Y\to X$ in distribution and each $X_n$ is integrable if and only if $Y$ is. – 2017-02-06

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