Using Dominated Convergence Theorem to prove that $\lim_{n\to\infty} \int_{\Lambda_n} X \, d\mathbb{P} = 0$ 
Let $\mathbb{E}\left(\left|X\right|\right) < \infty$ and $\lim_{n\to\infty} \mathbb{P}\left(\Lambda_n\right) = 0$, show that $\lim_{n\to\infty} \int_{\Lambda_n} X \, d\mathbb{P} = 0$.

I came up with the following proof, which does not use DCT. My friend said that using DCT the proposition is trivial to prove but did not offer further explanation. I honestly don't see how to use DCT for this proof. Do we have a shorter/easier proof using DCT?
My proof:
Since $\mathbb{E}\left( \left|X\right| \right) < \infty$,  $\sum_{n=1}^\infty \mathbb{P}\left( \left|X \right| \ge n \right)$ converges by theorem 3.2.1. This implies that $\mathbb{P}\left( \left|X\right| \ge n \text{ i.o. }\right) = 0$, which is equivalent to $\mathbb{P}\left( \left| X \right| < n \text{ a.s. } \right) = 1$. By definition, this means that 
$$ \mathbb{P}\left( \left\{\omega\mid \cup_{m=1}^\infty \cap_{n=m}^\infty \left|X(\omega)\right| < n \right\}\right) = 1 $$
This means that there exists a null set $N$, such that for all $\omega \in \Omega - N$, there exists some $K$ such that $\left|X(\omega)\right| < n$ for all $n \ge K$. In particular, $\left|X\right|$ is almost everywhere bounded by $K$.
Since $\lim_{n\to\infty} \mathbb{P}\left(\Lambda_n\right) = 0$, for any $\epsilon > 0$, there exists some $N(\epsilon)$ such that whenever $n\ge N(\epsilon)$, $\mathbb{P}\left(\Lambda_n\right) < \epsilon$ holds. So we have
\begin{align}
\left| \int_{\Lambda_n} X \,d \mathbb{P} \right| &\le \int_{\Lambda_n} \left| X \right| \,d\mathbb{P} \\
&\le K \mathbb{P}\left(\Lambda_n\right) \\
&\le K \epsilon
\end{align}
This proves that
$$ \lim_{n\to\infty} \int_{\Lambda_n} X \, d\mathbb{P} = 0 $$
Update: My original proof is incorrect as @Trevor Gunn pointed out, since I overlooked the fact that bound $K$ depends on $\omega$. The following proof is correct (perhaps unnecessarily verbose).
For any $K \in \mathbb{N}$, we have
\begin{align}
\left| \int_{\Lambda_n} X \,d\mathbb{P} \right| &\le \int_{\Lambda_n} \left| X \right| \,d\mathbb{P}\\
&= \int_{\Lambda_n} \left|X\right| 1_{\left|X\right| \le K} + \left|X\right| 1_{\left|X \right| > K} \, d\mathbb{P} \\
&\le \int_{\Lambda_n} \left| X1_{\left|X\right| \le K}\right| \,d\mathbb{P} + \int_{\Lambda_n} \left| X 1_{\left|X\right| > K} \right|\,d\mathbb{P} \\
&\le \int_{\Lambda_n} \left| X1_{\left|X\right| \le K}\right| \,d\mathbb{P} + \int_\Omega \left| X 1_{\left|X\right| > K} \right|\,d\mathbb{P}
\end{align}
Since $\left| X 1_{\left|X\right| \le K}\right| = \left|X\right| 1_{\left|X\right| \le K} \le K$ almost everywhere on $\Lambda_n$, we have
$$ \int_{\Lambda_n} \left| X 1_{\left|X\right|\le K} \right|\,d\mathbb{P} \le K \mathbb{P}\left( \Lambda_n \right) $$
In addition, notice that $\lim_{K\to\infty} \left| X 1_{\left|X\right| > K} \right| = 0$ almost everywhere on $\Omega$, as well as that $\left| X 1_{\left|X\right| < K}\right| \le \left|X \right|$ almost everywhere on $\Omega$. Since $\int_\Omega \left|X\right| \,d\mathbb{P} = 0$, by dominated convergence theorem, we conclude that
$$ \lim_{K\to\infty} \int_\Omega \left| X 1_{\left|X\right| > K}\right| \,d\mathbb{P} = 0 $$
Finally, given any $\epsilon > 0$, we can first choose $N \in \mathbb{N}$ such that $\int_\Omega \left| X 1_{\left|X\right| > K } \right| \,d \mathbb{P} < \frac{\epsilon}{2}$ whenever $K \ge N$. Having chosen $N$, choose $M$ such that $\mathbb{P}\left(\Lambda_n\right) < \frac{\epsilon}{2N}$ whenever $n\ge M$. Thus, when $n\ge M$, we have
$$\left| \int_{\Lambda_n} X  \, d\mathbb{P} \right| \le    N \frac{\epsilon}{2N} + \frac{\epsilon}{2} = \epsilon$$
This proves that
$$ \lim_{n\to\infty} \int_{\Lambda_n} X \,d\mathbb{P} = 0$$
 A: For $m\geq 0$, write  $$\int 1_{\Lambda_n}X dP = \int 1_{\Lambda_n}1_{|X|\leq m} X dP+\int 1_{\Lambda_n}1_{|X|> m} X dP$$
Then $\displaystyle \left|\int 1_{\Lambda_n}X dP\right|\leq \left|\int 1_{\Lambda_n}1_{|X|\leq m} X dP\right| + \left|\int 1_{\Lambda_n}1_{|X|> m} X dP\right|\leq mP(\Lambda_n) + \int 1_{|X|> m} |X| dP $
Now, let $\epsilon >0$. Since $|X|$ is finite a.e, by the DCT, $\int 1_{|X|> m} |X| dP$ converges to $0$ as $m\to \infty$.
Thus, there is some $m$ such that $\int 1_{|X|> m} |X| dP\leq \epsilon$. Furthermore, there is some $N$ such that $n\geq N \implies P(\Lambda_n)\leq \frac {\epsilon}{m}$.
For $n\geq N$, we have the bound $\displaystyle \left|\int 1_{\Lambda_n}X dP\right|\leq 2\epsilon$, hence the claim.

The simple approach below is flawed as $1_{\Lambda_n}$ needs not converge pointwise to $0$.
There's indeed a trivial proof resorting to DCT. Write $\int_{\Lambda_n} X \, d\mathbb{P}=\int 1_{\Lambda_n}X dP$.
Since $1_{\Lambda_n}$ converges pointwise to $0$, so does $1_{\Lambda_n}X$.
In addition, you have the trivial bound $|1_{\Lambda_n}X|\leq |X|$ with $\int |X| dP <\infty$, hence DCT applies and $$\lim_n \int 1_{\Lambda_n}X dP = \int 0 dP = 0$$
A: I agree with you. Consider the following example: take $[0,1]$ to be the measure space, $X = 1$, and $(\Lambda_n)$ to be the sequence that goes
$$ [0,1], [0,1/2], [1/2,1], [0,1/4], [1/4,2/4], [2/4,3/4], [3/4,1], [0,1/8], [1/8,2/8],\dots $$
and so on. Then clearly $\mathbf{P}(\Lambda_n) \to 0$ but there does not seem to be a way to make the indicator functions $I_{\Lambda_n}$ converge, which is what is needed to apply dominated convergence. I think your friend is assuming that $\lim_n I_{\Lambda_n} = 0$.
For your proof, you should also keep in mind that $K$ depends on $\omega$. Instead, as Gabriel does in his answer, split the integral into two pieces:
\begin{align}
\left\lvert \int_{\Lambda_n} X \,d\mathbf{P} \right\rvert &\le \left\lvert \int_{\Lambda_n} X I_{X \le K} \,d\mathbf{P} \right\rvert + \left\lvert \int_{\Lambda_n} X I_{X > K} \,d\mathbf{P} \right\rvert \\
&\le K \mathbf{P}(\Lambda_n) + \int |X| I_{X > K} \,d\mathbf{P}
\end{align}
and use the fact that $\mathbf{P}(\Lambda_n) \to 0$ for the first piece, and dominated convergence for the second.
A: Alternative, in my view more elementary.
It is enough to prove that $\lim_{n\to\infty}\int_{\Lambda_n}|X|dP\to0$.
Let $\epsilon>0$.
Let $g$ be a measurable function with finite image (so a step function) and $0\leq g\leq|X|$ and $\int gdP>\int |X|dP-\frac12\epsilon$. 
Then $g$ has the shape $\sum_{k=1}^ma_k1_{A_k}$ with $a_k\geq0$ and the $A_k$ measurable, disjoint and covering.
Then it is not difficult to prove that $\lim_{n\to\infty}\int g1_{\Lambda_n^{\complement}}dP=\int gdP$.
This because evidently $P(A_k\cap\Lambda_n^{\complement})\to P(A_k)$ for $k=1,\dots,m$.
So some $N$ exists such that $n>N$ implies that $\int g1_{\Lambda_n^{\complement}}dP>\int gdP-\frac12\epsilon$.
Then $\int |X|1_{\Lambda_n^{\complement}}dP\geq\int g1_{\Lambda_n^{\complement}}dP>\int gdP-\frac12\epsilon>\int |X|dP-\epsilon$ for $n>N$.
This proves that $\int |X|1_{\Lambda_n^{\complement}}dP$ converges to $\int|X|dP$ and consequently $\int |X|1_{\Lambda_n}dP$ converges to $0$.
