Conditions for convergence of moments Let ${X_n}$ be a sequence of r.v. such that $X_n\xrightarrow [d]{}X$, with $E(X)$ finite, and with $E(|X_n|^{1+\delta})\leq K<\infty$ for all $n$.
We know that:
a) For $\delta>0$, we have $E(|X_n|^{1+\delta})=(1+\delta)\int^{\infty}_0x^{\delta}P(|X_n|>x)dx$
b) For $M,\delta>0$, we have $\int^M_0P(|X_n|>x)dx\leq E(|X_n|)\leq \int^M_0P(|X_n|>x)dx+\int^{\infty}_M\left(\frac{x}{M}\right)^{\delta}P(|X_n|>x)dx$
Now I'm supposed to be able to prove that $E(|X_n|)\rightarrow E(|X|)$ and $E(X_n)\rightarrow E(X)$
I was trying to use $$|E(|X_n|)-E(|X|)|= \left|\int^M_0P(|X_n|>x)-P(|X|>x)dx + \int^{\infty}_M P(|X_n|>x)-P(|X|>x)dx\right|$$
$$\leq \underbrace{\int^M_0|P(|X_n|>x)-P(|X|>x)|dx }_{\rightarrow 0 \text{ as n goes to } \infty}+\left|\int^{\infty}_M P(|X_n|>x)-\int^{\infty}_MP(|X|>x)dx\right|$$
However, I'm not sure on how to deal with the 2nd term of this RHS and with b). Aren't each of the integrals finite? Couldn't I just apply $\lim_{n\rightarrow \infty}$? But then why would I need b)?
But otherwise, if I bounded it by $\int^{\infty}_M \left(\frac{x}{M}\right)^{\delta} |P(|X_n|>x)-P(|X|>x)|dx$, how would it help me?
Any help would be appreciated.
 A: First of all, note that it suffices to show that
$$\mathbb{E}(X_n) \to \mathbb{E}(X). \tag{1}$$
Then it follows from the triangle inequality that
$$|\mathbb{E}(|X_n|)-\mathbb{E}(|X|)| \leq \mathbb{E}(|X_n-X|) \to 0 \qquad \text{as} \, \, n \to \infty,$$
i.e. $\mathbb{E}(|X_n|) \to \mathbb{E}(|X|)$. So let's prove $(1)$. 
Fix $\varepsilon>0$. We have
$$\begin{align*} |\mathbb{E}(X_n)-\mathbb{E}(X)| &= \left| \int_0^{\infty} (\mathbb{P}(|X_n| \geq x) -\mathbb{P}(|X|>x)) \, dx \right| \\ &\leq \int_0^M |\mathbb{P}(|X_n| \geq x)-\mathbb{P}(|X| \geq x)| \, dx + \int_M^{\infty} \mathbb{P}(|X_n| \geq x) + \int_M^{\infty} \mathbb{P}(|X| \geq x) \, dx\\ &= I_1+I_2+I_3 \end{align*}$$
for any $M>1$. We estimate the terms separately. Since $\mathbb{E}(|X|)$ is finite, we can choose $M \geq M_1$ sufficiently large such that $I_3 \leq \frac{\varepsilon}{3}$. For the second term, note that
$$ \int_M^{\infty} \mathbb{P}(|X_n| \geq x) \leq \frac{1}{M^{\delta}} \int_0^{\infty} x^{\delta} \mathbb{P}(|X_n| \geq x) \, dx \leq \frac{K}{M^{\delta} (1+\delta)}.$$
This means that $I_2 \leq \frac{\varepsilon}{3}$ for $M \geq M_2$ sufficiently large. Now fix $M \geq \max(M_1,M_2)$. Since $X_n \to X$ in distribution, we have $\mathbb{P}(|X_n| \geq x) \to \mathbb{P}(|X| \geq x)$ for (Lebesgue-)almost all $x$. Hence, by the dominated convergence theorem,
$$\int_0^M |\mathbb{P}(|X_n| \geq x)-\mathbb{P}(|X| \geq x)| \, dx \to 0 \qquad \text{as}\,\, n \to \infty.$$
Consequently, we find
$$\limsup_{n \to \infty} |\mathbb{E}(X_n)-\mathbb{E}(X)| \leq 0 + \frac{\varepsilon}{3}+\frac{\varepsilon}{3} = \frac{2}{3} \varepsilon.$$
Since $\varepsilon>0$ was arbitrary, we conclude
$$\lim_{n \to \infty} |\mathbb{E}(X_n)-\mathbb{E}(X)| = 0.$$
