Convergence of $r$th moment implies uniform integrability (Chung Theorem 4.5.4.)

Question

Suppose $0<r<\infty$, $X_n\in L^r$, $X_n\overset{p}\to X$, and $E|X_n|^r\to E|X|^r<\infty$. We claim that $\{|X_n|^r\}$ is uniformly integrable.

To do so, Chung introduces a continuous function $f_A$ (for $A>0)$ so that $f_A(x)=|x|^r$ when $|x|^r\le A$, $f_A(x)\le|x|^r$ for $|x|^r\in(A,A+1]$, and $f_A(x)=0$ for $|x|^r>A+1$. We next note that

$\liminf_n\int_{|X_n|^r\le A+1}|X_n|^r\,dP\ge\lim_nE(f_A(X_n))=E(f_A(X))\ge\int_{|X|^r\le A}|X|^r\,dP$.

So far so good. Then he says "Subtracting from the limit relation in (iii), we obtain

$\limsup_n\int_{|X_n|^r>A+1}|X_n|^r\,dP\le\int_{|X|^r>A}|X|^r\,dP$", where (iii) refers to $\lim_n E(|X_n|^r)=E(|X|^r)<\infty$.

I am having trouble making this leap, and would greatly appreciate it if someone could help me see how we get this inequality.

Thanks in advance.

Answer 1

Here's a hint: Take the inequality: $$\liminf_t \int_{|X_n|^r \leq A + 1} |X_n|^r \,dP \geq \int_{|X|^r \leq A} |X|^r \,dP$$ and negate both sides to get: $$\limsup_t \int_{|X_n|^r \leq A + 1} -|X_n|^r \,dP \leq \int_{|X|^r \leq A} -|X|^r \,dP.$$ What happens when you add this to both sides of (iii)?

Answer 2

For the moment, we have $$\liminf_n\int_{|X_n|^r\leqslant A+1}|X_n|^r\,dP\geqslant\int_{|X|^r\leqslant A}|X|^r\,dP,$$ hence $$\limsup_n -\int_{|X_n|^r\leqslant A+1}|X_n|^r\,dP\leqslant -\int_{|X|^r\leqslant A}|X|^r\,dP.$$

Fact: if $\left(x_n\right)_{n\geqslant 1}$ and $\left(y_n\right)_{n\geqslant 1}$ are two sequence of real numbers such that $y_n\to y\in\mathbb R$, then $\limsup_{n\to +\infty}x_n+y_n=y+ \limsup_{n\to +\infty}x_n$.

Use this with $x_n:= -\int_{|X_n|^r\leqslant A+1}|X_n|^r\,dP$, $y_n=\int |X_n|^r\,dP$ and $y=\int |X|^r\,dP$, we get the wanted result.