$\mathbb E[|X_n|^r]<\infty$ and $\mathbb E[|X_n|^r]\to \mathbb E[|X|^r]$ as $n\to \infty$

Question

Let $\{X_n\}\xrightarrow{d}X$ and for some $p>0$, we have $$\sup_{n\ge 1} \mathbb E[|X_n|^p]<\infty$$ Show that for any $r\in (0,p)$, we have

a. $\mathbb E[|X|^r]<\infty$

b. $\mathbb E[|X_n|^r]\to \mathbb E[|X|^r]$ as $n\to \infty$

[Note: You must not use (b) to prove (a)]

The actual question had $\mathbb E[|X|^r]<\infty$ instead of $\mathbb E[|X_n|^r]<\infty$ which was wrongly written in the first question. So, now I have doubts in part (a) as well. The $1\le r\le p$ case can be tackled using some theorems done in class, but I can't find any argument for the $0<r\le 1$ case.

I am pretty sure we need to use Skorohod Representation Theorem to get $\{Y_n\}$ and $Y$ which are equal to $\{X_n\}$ and $X$ in distribution and $Y_n\xrightarrow{a.s.} Y$.

Maybe, we also need to use the fact that $$\{X_n\}\xrightarrow{d}X \iff \mathbb E[f(X_n)]\to \mathbb E[f(X)]\;\;\forall f\in \mathcal C_B(\mathbb R)$$ but I can't figure out how to do that since the map $x\mapsto x^r$ is not bounded.

Answer 1

Since $Y_n$ and $X_n$ have the same distribtuion function we have $E|X_n|^{p}=E|Y_n|^{p}$. So $\sup_n E|Y_n|^{p}<\infty$. Let $Z_n=|Y_n|^{r}$. Then $\sup E|Z_n|^{p/r} <\infty$. This makes $(Z_n)$ uniformly integrable (because $\frac p r >1$). Hence, $EZ_n \to EZ$ where $Z=|Y|^{r}$.