How is this done by dominated convergence theorem? Terry Tao wrote in his blog

Fix ${k \geq 1}$. If ${X}$ has finite ${k^{th}}$ moment, say ${{\bf E}|X|^k \leq C}$, then from Markov’s inequality (14) one has
  $$
\displaystyle {\bf P}(|X| \geq \lambda) \leq C \lambda^{-k}, \ \ \ \ \ (23)
$$
  thus we see that the higher the moments that we control, the faster the tail decay is. From the dominated convergence theorem we also have the variant
  $$
\displaystyle \lim_{\lambda \rightarrow \infty} \lambda^k {\bf P}(|X| \geq \lambda) = 0. \ \ \ \ \ (24)$$

I was wondering how (24) is derived from (23) and DCT?
Thanks!
 A: Well, I'm not entirely sure what he had in mind, but here is how I would prove it: first define the sets 
$$
A_\lambda=\{x\in\Omega:\vert X(x)\vert\geq\lambda\}
$$ Then, since $\vert X\vert^k\geq \lambda^k$ on the set $A_\lambda$, we have the following inequality: 
$$
\int_{A_\lambda}\vert X\vert^kd\textbf{P}\geq\lambda^k\textbf{P}(A_\lambda)
$$ This will be more useful than the inequality (23).  Since we are assuming that $X$ has a finite $k$th moment, 
$$
\lim_{\lambda\rightarrow\infty}\int_{A_\lambda}\vert X\vert^kd\textbf{P}=0
$$
This is where dominated convergence is needed (robjohn's answer makes this step clear).  We're basically done, but just to summarize: $0\leq\lambda^k\textbf{P}(A_\lambda)\leq\int_{A_\lambda}\vert X\vert^kd\textbf{P}$ and the latter converges to zero as $\lambda\rightarrow\infty$.  Hence by the squeeze theorem, the former does as well.
A: If we stop the proof of Markov's inequality short, we have
$$
\begin{align}
\mathrm{Pr(|X|\ge\lambda)}
&=\int_{|X|\ge\lambda}\,\mathrm{d}\mu\\
&\le\lambda^{-k}\int_{|X|\ge\lambda}|X|^k\,\mathrm{d}\mu
\end{align}
$$
This means that
$$
\begin{align}
\lambda^k\,\mathrm{Pr(|X|\ge\lambda)}
&\le\int_{|X|\ge\lambda}|X|^k\,\mathrm{d}\mu\\
&=\int{\Large\chi}_{|X|\ge\lambda}|X|^k\,\mathrm{d}\mu\\
\end{align}
$$
and since ${\Large\chi}_{|X|\ge\lambda}|X|^k\le|X|^k$ and $\lim\limits_{\lambda\to\infty}{\Large\chi}_{|X|\ge\lambda}|X|^k=0$ pointwise, DCT guarantees that
$$
\lim_{\lambda\to\infty}\int{\Large\chi}_{|X|\ge\lambda}|X|^k\,\mathrm{d}\mu=0
$$
Therefore,
$$
\lim_{\lambda\to\infty}\lambda^k\,\mathrm{Pr(|X|\ge\lambda)}=0
$$
