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Let $X_{n}$ be a sequence of random variables that converges to the random variable $X$ almost surely. I would like to show that if $$\sup_{n}{\mathbb{E}(|X_{n}|^{p})} \leq \infty$$

for $p>1$ then $$\mathbb{E}(|X_{n}-X|) \to 0$$

On the first glance, this sort of problem requires an indirect application of Lebesgue dominated convergence theorem but it is not clear enough how to use the finiteness of the moments.

On the other hand, i can show that if $X_{n} \to X$ almost surely and $\mathbb{E}{X_{n}} \to \mathbb{E}{X}$ then $\mathbb{E}|X_{n}-X| \to 0$.

Is it possible to derive the last statement from the first one and, if not, what are the possible ways to prove initial one?

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    The boundedness in $L^p$ implies that $(X_n)_n$ is uniformly integrable, and therefore the claim follows from Vitali's convergence theorem. – 2017-02-09
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    @saz Thanks, i have successfuly recovered the omitted detailes – 2017-02-09

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As @saz pointed out, the statement can be proven as follows:

First i recall that there exists a characterisation of a uniform continuity in terms of test functions. More precisely, a family of random variables $A = \{ X_{n} \}$ is uniformly integrable if and only if there exists a uniformly integrable function $\varphi$ so that $$\sup_{X \in A}{\mathbb{E}(\varphi(|X|)) < \infty}$$

Using the fact above we obtain the condition in which $$\sup_{n}{\mathbb{E}(|X_{n}|^{p})} < \infty$$ as a corollary. Then, the Vitali convergence theorem implies the result.