$\mathbf{E} |\xi_n - \xi|^a \to 0 \Rightarrow \mathbf{E} \xi_n^a \to \mathbf{E} \xi^a$. 
Let $\xi_n$, $\xi$ be nonnegative random variables. Prove or disprove that if
$$\mathbf{E} \lvert \xi_n - \xi \rvert^a \to 0 \quad \text{as } n\to \infty$$
for all $a \in (0, \infty)$, then
$$\mathbf{E} \xi_n^a \to   \mathbf{E} \xi^a$$

I know that if $a \in \mathbb{N}$ then the statement is true, but I don't know how to prove it and I don't know anything about the case when $a$ is arbitrary. The following is my attempt.
If $a=m \in \mathbb{N}$ we have
\begin{align}
\lvert \mathbf{E} \xi_n^m - \mathbf{E}  \xi^m \rvert &= \lvert \mathbf{E} (\xi_n - \xi)(\xi_n^{m-1} - \xi_n^{m-2}\xi + \ldots ) \rvert \\
&\le \sqrt{\mathbf{E} (\xi_n - \xi)^2 \mathbf{E}(\xi_n^{m-1} - \xi_n^{m-2}\xi + \ldots )^2} \\
&\to 0
\end{align}
because $\mathbf{E}(\xi_n^{m-1} - \xi_n^{m-2}\xi + \ldots )^2$ is bounded. It is bounded because it is a finite sum of terms $\mathbf{E}\xi_n^i \xi^j$ and
$$\lvert \mathbf{E}\xi_n^i \xi^j \rvert \le \sqrt{\mathbf{E}\xi_n^{2i} \mathbf{E}\xi^{2j} }$$
So it's sufficient to show that $\mathbf{E}\xi_n^{2i}$ is bounded for every $i$.
 A: Suppose $a \in (0, \infty)$ and $E(|\xi_n - \xi|^a) \to 0$ as $n \to \infty$. I will also assume that $\xi_n \in L^a$ for all $n \geq 1$ and $\xi \in L^a$, meaning that $E(|\xi_n|^a) < \infty$ for all $n \geq 1$ and $E(|\xi|^a) < \infty$. First suppose $a \geq 1$. Since $E(|\xi_n - \xi|^a) = \|\xi_n - \xi\|_{L^a}^a$, this says that $\xi_n \to \xi$ in $L^a$. Since $L^a$ is a normed vector space (the triangle inequality for this space is called Minkowski's inequality) with the norm $\|\cdot\|_{L^a}$ and the distance function on a metric space is continuous, we get $\|\xi_n\|_{L^a} \to \|\xi\|_{L^a}$. Taking $a$th powers yields the result.
Now suppose $0 < a < 1$. In this case, the inequality $|x + y|^a \leq x^a + y^a$ for $x, y \geq 0$ implies that the function $d : L^a \times L^a \to [0, \infty)$ defined by
$$d_{L^a}(f, g) = E(|f - g|^a)$$
makes $(L^a, d_{L^a})$ a metric space and by hypothesis, $\xi_n \to \xi$ in this metric. Since $d_{L^a}$ is continuous, we get $d_{L^a}(\xi_n, 0) \to d_{L^a}(\xi, 0)$, i.e.
$$E(|\xi_n|^a) \to E(|\xi|^a).$$
