If $\sup_n$ $E|X_n|^{1+\sigma} \lt \infty$ for $\sigma \gt $0, then $\{|X_n|\}$ is uniformly integrable.
I've seen a similar problem without the exponent on $|X_n|^{1+\sigma}$ and tried to apply it here but I think I might be missing something.
My attempt:
I started by defining $|X_n|^{1+\sigma} \leq Y $
Then
$$
\{|X_N|^{1+\sigma} \gt R\} \subseteq \{Y \gt R\}
$$
Therefore
$$
\int_{|X_N|^{1+\sigma} \gt R} |X_N|^{1+\sigma}\,dP \leq \int_{|X_N|^{1+\sigma} \gt R} Y \,dP \leq \int_{Y \gt R} |Y| \,dP
$$
Implying
$$
\sup\int_{|X_N|^{1+\sigma} \gt R} |X_N|^{1+\sigma}\,dP \leq \int_{Y \gt R} |Y| \,dP
$$
Does following this pattern make sense? Am I missing something? Any help would be appreciated.
Best Answer
The following is a well-known fact.
Proposition: Let $(\Omega,\mathcal{F},P)$ be a probability space. Let $\mathcal{C}=\{X_{i}\mid i\in\Lambda\}$ be a family of random variables (that family may be uncountable). If there exists $p\in(1,\infty)$ such that $$ \sup_{i\in\Lambda}\int|X_{i}|^{p}\,dP<\infty, $$ then $\mathcal{C}$ is uniformly integrable.
Proof: Let $l=\sup_{i\in\Lambda}\int|X_{i}|^{p}\,dP$. Let $q\in(1,\infty)$ be such that $\frac{1}{p}+\frac{1}{q}=1$. Let $\varepsilon>0$ be arbitrary. Choose $c>0$ such that $l^{\frac{1}{p}}\cdot\left(\frac{l}{c^{p}}\right)^{\frac{1}{q}}<\varepsilon$, which is possible because $l^{\frac{1}{p}}\cdot\left(\frac{l}{t^{p}}\right)^{\frac{1}{q}}\rightarrow0$ as $t\rightarrow\infty$. Define $A_{i}=\{\omega\in\Omega\mid|X_{i}(\omega)|\geq c\}$. Observe that \begin{eqnarray*} l & \geq & \int_{A_{i}}|X_{i}|^{p}\,dP\\ & \geq & P(A_{i})c^{p}. \end{eqnarray*} That is, $P(A_{i})\leq\frac{l}{c^{p}}.$ On the other hand, by Holder inequality, \begin{eqnarray*} \int_{A_{i}}|X_{i}|\,dP & = & \int|X_{i}|\cdot 1_{A_{i}}\,dP\\ & \leq & ||X_{i}||_{p}||1_{A_{i}}||_{q}\\ & \leq & l^{\frac{1}{p}}\left\{ P(A_{i})\right\} ^{\frac{1}{q}}\\ & \leq & l^{\frac{1}{p}}\left(\frac{l}{c^{p}}\right)^{\frac{1}{q}}\\ & < & \varepsilon. \end{eqnarray*} Therefore, $$ \sup_{i\in\Lambda}\int_{A_{i}}|X_{i}|\,dP\leq\varepsilon $$ and hence $\mathcal{C}$ is uniformly integrable.