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
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?