Suppose that each $X_n$ and $X$ are non-negative random variables and $X_n \stackrel d \to X.$ Assume that $\{X_n \}_{n \geq 1}$ is uniformly integrable. Prove the following decomposition $$\mathbb E(X_n) = \int_{0}^{M} \mathbb P (M \geq X_n \gt t)\ dt + \int_{X_n \gt M} X_n\ d \mathbb P.$$ Hence, show that $\mathbb E(X_n)$ converges to $\mathbb E(X).$ Argue that the same is true even if we did not assume non-negativity.
I am able to show that decomposition. I also showed that if $\{X_n\}_{n \geq 1}$ is a sequence of random variables (not necessarily non-negative) and $X_n \stackrel d \to X$ then $X_n^+ \stackrel d \to X^+$ and $X_n^- \stackrel d \to X^-.$ I also showed that if $\{X_n \}_{n \geq 1}$ is uniformly integrable then so are $\{X_n^+ \}_{n \geq 1}$ and $\{X_n^- \}_{n \geq 1}.$ So if we can able to show that $\mathbb E(X_n) \to \mathbb E(X)$ for non-negative sequence of random variables $\{X_n \}_{n \geq 1}$ satisfying uniform integrability I can able to extend the same idea for any sequence of random variables satisfying uniform integrability.
Could anyone kindly give some idea on how to use the decomposition to conclude that $\mathbb E(X_n) \xrightarrow {n \to \infty} \mathbb E(X)\ $? Thanks and Regards.
Recall that one of the definitions of uniform integrability (people use a lot of different ones, so if you have one in mind you have to specify) is that for any $\epsilon > 0$, there is some $M$ such that $$\mathbb E[|X_n| \cdot 1_{|X_n| > M}] = \int_{|X_n| > M} |X_n| d\mathbb P < \epsilon.$$
As you said, we may take $X_n$ nonnegative without loss of generality, so that we can drop the absolute values above. We first show that $\mathbb E[X] < \infty$, as by Fatou $$\mathbb E[X] = \int_0^\infty \lim_n \mathbb P(X_n > t)dt \leq \liminf_n \int_0^\infty \mathbb P(X_n > t)dt = \liminf_n \mathbb E[X_n] < \infty.$$
Now, by the decomposition in the problem, $$ \mathbb E[X] - \mathbb E[X_n] = \int_0^M (\mathbb P(t < X \leq M) - \mathbb P(t < X_n \leq M))dt + \int_{X>M} X d\mathbb P - \int_{X_n>M} X_n d\mathbb P. $$ For any $\epsilon > 0$, by uniform integrability and the fact that $\mathbb E[X] < \infty$, we can choose $M$ such that both of the latter integrals are bounded by $\epsilon$. Then the above reduces to $$ \mathbb E[X] - \mathbb E[X_n] \leq \int_0^M (\mathbb P(t < X \leq M) - \mathbb P(t < X_n \leq M))dt + 2\epsilon. $$ Now, by convergence in distribution, $$\mathbb P(t < X \leq M) - \mathbb P(t < X_n \leq M) \to 0$$ and we can conclude by dominated convergence that for sufficiently large $n$, $$\int_0^M (\mathbb P(t < X \leq M) - \mathbb P(t < X_n \leq M))dt < \epsilon.$$
So we have shown that for any $\epsilon > 0$, for sufficiently large $n$, $$\mathbb E[X] - \mathbb E[X_n] < 3\epsilon$$ and so we may conclude.