Convergence of Moments (3)

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Let $\{X_n\}$ and $X$ be random variables.  Let $0<r<\infty$, $X_n\in L^r$, and $X_n\rightarrow X$ in probability.  Then the following three propositions are equivalent.

(1) $\{|X_n|^r\}$ is uniformly integrable;
(2) $X_n\rightarrow X$ in $L^r$;
(3) $\mathscr{E}|X_n|^r\rightarrow\mathscr{E}|X|^r<\infty$.

$\bullet$ Proof.

(1) $\implies$ (2):
Suppose (1) is true, we have [See details]
(a) $\mathscr{E}|X_n|^r\leq M<\infty$ for all $n$.
(b) For every $\varepsilon>0$, there exists $\delta(\varepsilon)>0$ such that for any $E\in\mathscr{F}$, $$\mathscr{P}(E)<\delta(\varepsilon)\implies\int_E|X_n|^r\,d\mathscr{P}<\frac{\varepsilon}{2}\mbox{ for every }n.$$Additionally, since $X_n\overset{p}{\longrightarrow}X$, we have that there exists a subsequence $\{n_k\}$ such that $X_{n_k}\overset{a.s.}{\longrightarrow}X$.  Combining the (a) and $X_{n_k}\overset{a.s.}{\longrightarrow}X$, by the first part of Convergence of Moments (1), we have $$\mathscr{E}|X|^r\leq\underset{n\rightarrow\infty}{\underline{\lim}}\mathscr{E}|X_n|^r\leq\underset{n\rightarrow\infty}{\overline{\lim}}\mathscr{E}|X_n|^r\leq M<\infty.$$Which means $X\in L^r$, i.e. $$\forall\,\varepsilon>0,\,\exists\,\delta(\varepsilon)>0\mbox{ s.t. }\forall\,E'\in\mathscr{F},\,\mathscr{P}(E')<\delta\implies\int_{E'}|X|\,d\mathscr{P}<\frac{\varepsilon}{2}.$$  Next, for each $r>0$, by the following inequality [See detials] $$\mathscr{E}|X_n-X|^r\leq 2^r\{\mathscr{E}|X_n|^r+\mathscr{E}|X|^r\},$$we have
(a') $\mathscr{E}|X_n-X|^r\leq2^r(M+M)<\infty$.
(b') Choose $E_0$ to be the event with smaller probability among $\mathscr{P}(E)$ and $\mathscr{P}(E')$, we have $$\int_{E_0}|X_n-X|^r\,d\mathscr{P}\leq\int_{E_0}|X_n|^r\,d\mathscr{P}+\int_{E_0}|X|^r\,d\mathscr{P}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}.$$Thus, $\{X_n-X\}$ is also uniformly integrable.  Finally, for every $\varepsilon>0$, $$\begin{array}{rl}\int_\Omega|X_n-X|^r\,d\mathscr{P}&=\int_{|X_n-X|>\varepsilon}|X_n-X|^r\,d\mathscr{P}+\int_{|X_n-X|\leq\varepsilon}|X_n-X|^r\,d\mathscr{P}\\ &\leq\int_{|X_n-X|>\varepsilon}|X_n-X|^r\,d\mathscr{P}+\varepsilon^r\rightarrow0 \end{array}$$since $\mathscr{P}\{|X_n-X|>\varepsilon\}\rightarrow0\implies\int_{|X_n-X|>\varepsilon}|X_n-X|^r\,d\mathscr{P}\rightarrow0$ by the uniformly integrability.  We have $X_n\overset{L^r}{\longrightarrow}X$.

(2) $\implies$ (3):
Please see the second part of Convergence of Moments (1).

(3) $\implies$ (1):
Let $A>0$ and define $$f_A(x)\begin{cases}=|x|^r,&\mbox{for }|x|^r\leq A;\\ \leq|x|^r,&\mbox{for }A<|x|^r\leq A+1;\\ =0,&\mbox{for }|x|^r>A+1. \end{cases}$$Then $f_A\in C_K$ and by Converge in Distribution and Vague Convergence (1): Equivalence for s.p.m.'s, we have $$\underset{n\rightarrow\infty}{\underline{\lim}}\int_{|X_n|^r\leq A+1}|X_n|^r\,d\mathscr{P}\geq\underset{n\rightarrow\infty}{\lim}\mathscr{E}\left\{f_A(X_n)\right\}=\mathscr{E}\left\{f_A(X)\right\}\geq\int_{|X|^r\leq A}|X|^r\,d\mathscr{P},$$because $X_n\overset{p}{\longrightarrow}X\implies X_n\overset{d}{\longrightarrow}X$.  Then we have $$\underset{n\rightarrow\infty}{\overline{\lim}}\int_{|X_n|^r>A+1}|X_n|^r\,d\mathscr{P}\leq\int_{|X|^r> A}|X|^r\,d\mathscr{P}\rightarrow0\mbox{  as }A\rightarrow\infty.$$Which means for every $\varepsilon>0$, there exists $A_0=A_0(\varepsilon)$ and $n_0=n_0(A_0(\varepsilon))$ such that $$\underset{n>n_0}{\sup}\int_{|X_n|^r>A+1}|X_n|^r\,d\mathscr{P}<\varepsilon$$for $A>A_0$.  Since each $X_n$ is integrable, so we can choose $A_1$ such that $$\underset{n\geq1}{\sup}\int_{|X_n|^r>A_1+1}|X_n|^r\,d\mathscr{P}<\varepsilon.$$Hence, let $A=\max{\{A_0, A_1\}}+1$, we have $$\int_{|X_n|^r>A}|X_n|^r\,d\mathscr{P}<\varepsilon\mbox{  for all }n,$$that is $\{X_n\}$ is uniformly integrable.

$\Box$