Uniformly integrable for a random variable I am trying to show that for $k&gt;0$ and $X$ a real valued r.v. with $E[| X|^k]&lt;\infty$, we have:
$$
\lim_{M\to\infty} E[|X|^k 1_{\{|X|&gt;M\}}]=0
$$

I was trying to apply Holder's inequality with $p=1,q=\infty$ to the expectation to separate it into two terms, then use Markov's inequality to get a finite expectation divided by M. However, I don't think I understand the case $q=\infty$ very well.

Question: is there another way to solve this without Holder?

If you have any tips on how to solve this, please comment below. Thanks for helping! :D
Consider the partition:
$$
\begin{align}
&amp;\Omega=\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|&gt;M\}}=0\right\}\cup\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|&gt;M\}}\neq0\right\}\\
\end{align}
$$
Consider the set to right. A necessary condition for the limit inside it to be different than zero is that the indicator function is different than zero for infinitely many values of $M$. Since $M\to\infty$ and $X:\Omega\to\bar{\mathbb{R}}$, we must have $X(\omega)&gt;M,\forall M$, which implies that $X(\omega)=\infty$. Thus:
$$
\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|&gt;M\}}\neq0\right\}\subset\left\{\omega\in\Omega:X(\omega)=\infty\right\}
$$
Now, if the size of that latter set is greater than zero:
$$
\begin{align}
&amp;P\left(\left\{\omega\in\Omega:X(\omega)=\infty\right\}\right)&gt;0\\
&amp;\implies\mathrm{E}\left(\left|X\right|^\frac{k}{2}\right)\geq \left|X(\omega)\right|^\frac{k}{2}P\left(\left\{\omega\in\Omega:X(\omega)=\infty\right\}\right)=\infty
\end{align}
$$
Which contradicts the hypothesis $\mathrm{E}\left(\left|X\right|^\frac{k}{2}\right)=\infty$. So:
$$
\begin{align}
&amp;P\left(\left\{\omega\in\Omega:X(\omega)=\infty\right\}\right)=0\\
\end{align}
$$
Now using the first equation (partition):
$$
\begin{align}
1&amp;=P\left(\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|&gt;M\}}=0\right\}\right)+P\left(\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|&gt;M\}}\neq0\right\}\right)\\
&amp;\leq P\left(\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|&gt;M\}}=0\right\}\right)+P\left(\left\{\omega\in\Omega:X(\omega)=\infty\right\}\right)\\
&amp;\leq P\left(\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|&gt;M\}}=0\right\}\right)
\end{align}
$$
So we have that:
$$
\begin{align}
\left| X\right|^k 1_{\{\left |X\right|&gt;M\}}\overset{a.s.}{\to}0
\end{align}
$$
Since that sequence (in $M$) is bounded by $\left|X\right|^k$, which is bounded in expectation, the Dominated Convergence theorem gives the desired result.
Consider the partition:
$$
\begin{align}
&amp;\Omega=\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|&gt;M\}}=0\right\}\cup\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|&gt;M\}}\neq0\right\}\\
\end{align}
$$
Consider the set to right. A necessary condition for the limit inside it to be different than zero is that the indicator function is different than zero for infinitely many values of $M$. Since $M\to\infty$ and $X:\Omega\to\bar{\mathbb{R}}$, we must have $X(\omega)&gt;M,\forall M$, which implies that $X(\omega)=\infty$. Thus:
$$
\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|&gt;M\}}\neq0\right\}\subset\left\{\omega\in\Omega:X(\omega)=\infty\right\}
$$
Now, if the size of that latter set is greater than zero:
$$
\begin{align}
&amp;P\left(\left\{\omega\in\Omega:X(\omega)=\infty\right\}\right)&gt;0\\
&amp;\implies\mathrm{E}\left(\left|X\right|^\frac{k}{2}\right)\geq \left|X(\omega)\right|^\frac{k}{2}P\left(\left\{\omega\in\Omega:X(\omega)=\infty\right\}\right)=\infty
\end{align}
$$
Which contradicts the hypothesis $\mathrm{E}\left(\left|X\right|^\frac{k}{2}\right)=\infty$. So:
$$
\begin{align}
&amp;P\left(\left\{\omega\in\Omega:X(\omega)=\infty\right\}\right)=0\\
\end{align}
$$
Now using the first equation (partition):
$$
\begin{align}
1&amp;=P\left(\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|&gt;M\}}=0\right\}\right)+P\left(\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|&gt;M\}}\neq0\right\}\right)\\
&amp;\leq P\left(\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|&gt;M\}}=0\right\}\right)+P\left(\left\{\omega\in\Omega:X(\omega)=\infty\right\}\right)\\
&amp;\leq P\left(\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|&gt;M\}}=0\right\}\right)
\end{align}
$$
So we have that:
$$
\begin{align}
\left| X\right|^k 1_{\{\left |X\right|&gt;M\}}\overset{a.s.}{\to}0
\end{align}
$$
Since that sequence (in $M$) is bounded by $\left|X\right|^k$, which is bounded in expectation, the Dominated Convergence theorem gives the desired result.