Proof of a lemma in probability I've encountered this lemma in Chung's book as an exercise:

If $\mathbb{E}|X|<\infty$ and $
 \lim_{n \to \infty}
 \mathbb{P}\{\Lambda_{n}\} = 0$,
  then, $$\lim_{n \to \infty}
 \int_{\Lambda_{n}}
 X\,\mathrm{d}\mathbb{P} = 0 \>.$$

Could anyone provide a detailed proof?
I'm wondering since $\mathbb{E}|X|<\infty$, can I use the fact $|X|<\infty \;\mathrm{a.e.}$ then $\exists M \in \mathbb{R}^{+} \,\mathrm{s.t.}\, |X|<M \;\mathrm{a.e.}$
Then $\int_{\Lambda_{n}} X\,\mathrm{d}\mathbb{P} \leq M \,\mathbb{P}\{\Lambda_{n}\}\rightarrow 0$.
And, can I use this lemma to prove that every $X \in L^{1}$ is uniformly integrable, using Thm 4.5.3 in Chung's book 'A course in probability theory'?
Hence, every finite set of $\{X_{n} \subset L^{1}\}$ is uniformly integrable.
However, why infinite set (possibly countably infinite) may not be uniformly integrable?
Sorry to entangle these two questions together.
 A: Hint: Write $\int_{\Lambda_n} X d\mathbb{P}$ as $\int X 1_{\Lambda_n} d\mathbb{P}$, and note that $|X 1_{\Lambda_n}| \le |X|$.  Then a certain familiar theorem applies...
A: See Propostion 4.16 and its proof here. The result you are interested in is an immediate corollary.
EDIT: Adapting that proposition to our setting, it can be stated as follows: Suppose that $X: \Omega \to \mathbb{R}$ is an integrable random variable, meaning that $
\int {|X|{\rm d\mathbb P} }  < \infty $. Then, given any $\varepsilon > 0$, there exists $\delta > 0$ such that
$$
0 \le \int_A {|X|{\rm d\mathbb P} }  < \varepsilon 
$$
whenever $A$ is a measurable set with ${\mathbb P}(A)< \delta$. (Hence if ${\mathbb P}(\Lambda _n) \to 0$, we conclude that $\lim _{n \to \infty } \int_{\Lambda _n } {X{\rm d\mathbb P}}  = 0$.)
A: Let $X_n :=|X|\mathbb{1}_{\left\{|X|\leq n\right\}}$. Since each $X_n$ is integrable and $X$ is finite almost everywhere we have from the Lebesgue monotone convergence theorem that $\displaystyle \lim_{n\to +\infty} \int_{\Omega}X_nd\mathbb{P} = \int_{\Omega} Xd\mathbb{P}$.
Let $\varepsilon >0$. We can find $n_0$ such that $\int_{\Omega}|X|\mathbb{1}_{\left\{|X|\geq n_0\right\}}d\mathbb{P}<\frac{\varepsilon}2$. We have
$$
\left|\int_{\Lambda_n}Xd\mathbb{P}\right|\leq \left|\int_{\Lambda_n}X\cdot \mathbb{1}_{\left\{|X|\geq n_0\right\}}d\mathbb{P}\right| +\left|\int_{\Lambda_n}X\cdot \mathbb{1}_{\left\{|X|< n_0\right\}}d\mathbb{P}\right|\leq 
\int_{\Omega} |X|\mathbb{1}_{\left\{|X|\geq n_0\right\}}d\mathbb{P} +n_0P(\Lambda_n)
$$
and we can conclude taking $N$ such that if $n\geq N$ then $P(\Lambda_n)\leq \frac{\varepsilon}{2n_0}$.
A: I have five words for you: absolute continuity of the integral. :)
