Convergence of moments and absolute moments of random variables Let $X$ and $(X_n)_{n\geq 1}$ be random variables such that $X_n\to X$ in distribution. Assume that $\sup_n E[|X_n|^r]<\infty$ for some $r>0$. Then how do I show that $E[|X|^r]<\infty$ and that $E[X_n^\alpha]\to E[X^\alpha]$ and $E[|X_n|^\alpha]\to E[|X|^\alpha]$ for all $0<\alpha < r$. I'm told to show and use that $\sup_n E[|X_n|^\alpha - |X_n|^\alpha \wedge m]\to 0$ for $m\to\infty$ for all $\alpha<r$, but I don't see how this helps me. Any thoughts?
My thoughts on the first is the following:
$$
E[|X|^r]=\sup_{k\in\mathbb{N}}E[|X|^r\wedge k]=\sup_{k\in\mathbb{N}}\lim_{n\to\infty}E[|X_n|^r\wedge k]\leq \sup_{k\in\mathbb{N}}\sup_{n\in\mathbb{N}}E[|X_n|^r\wedge k]\leq \sup_{n\in\mathbb{N}}E[|X_n|^r]<\infty
$$
Is this correct?
 A: We have using Hölder's inequality and denoting $M:=\sup_{n\geq 1}E|X_n|^r$ 
\begin{align*}
\int |X_n|^{\alpha}\mathbf 1_{|X_n|^{\alpha}> m}&\leq \left(\int |X_n|^r\right)^{\alpha/r}\left(P(|X_n|^\alpha> m)\right)^{\frac {r-\alpha}r}\\
&\leq M^{\alpha /r}\left(P(|X_n|^r> m^{\frac r{\alpha}})\right)^{\frac {r-\alpha}r}\\
&\leq M^{\alpha/r}\left(\frac 1{m^{\frac r{\alpha}}}E |X_n|^r\right)^{\frac {r-\alpha}r}\\
&=M\frac 1{m^{\frac{r-\alpha}{\alpha}}},
\end{align*}
and since $\frac{r-\alpha}{\alpha}>0$ we have the wanted convergence. 
Now, fix $\varepsilon>0$ and pick an integer $m_0$ such that $E|X|^{\alpha}\wedge m_0\leq \varepsilon$ and $\sup_n E|X_n|^{\alpha}\wedge m_0\leq \varepsilon$. 
Then for all $n_0$
$$\sup_{n\geq n_0}|E|X_n|^{\alpha}-E|X|^{\alpha}|\leq \varepsilon+\sup_{n\geq n_0}|E|X_n|^{\alpha}\wedge m_0-E|X|^{\alpha}|,$$
and using the fact that the map $x\mapsto |x|^{\alpha}\wedge m$ is continuous and bounded we take $n_0$ such that $\sup_{n\geq n_0} |E|X_n|^{\alpha}\wedge m_0-E|X|^{\alpha}\wedge m_0|\leq \varepsilon$ and we can conclude that $\sup_{n\geq n_0}|E|X_n|^{\alpha}-E|X|^{\alpha}|\leq 3\varepsilon$.
