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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.

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The fact that $|X| \lt \infty$ a.e. does not imply the existence of an $M$ with $|X| \lt M$ a.e. Consider $1/x^{1/2}$, on $[0,1]$ for example. –  t.b. Jun 19 '11 at 13:57
    
Since you insist on uniform integrabiliby: isn't it obvious that $f_n = 1_{\Lambda_n} X$ is uniformly integrable, as $|f_n| \leq |X|$? –  t.b. Jun 19 '11 at 14:14
    
@Theo Buehler: hmm, it seems I had a mistake at the beginning. Could you give me any hints on this? Thank you. –  newbie Jun 19 '11 at 14:15
    
I think uniform integrability is settled. Clearly $f_n \to 0$ in measure. –  t.b. Jun 19 '11 at 14:18
    
Yes, it is true. But can the lemma be proved without introducing uniform integrability? –  newbie Jun 19 '11 at 14:20

4 Answers 4

up vote 3 down vote accepted

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...

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yes, dominated convergence theorem can be appied here. I know it sounds stupid, but I couldn't give a logical reasoning for $X1_{\Lambda_{n}} \rightarrow 0$ in measure from $\mathbb{P}\{\Lambda_{n}\} \rightarrow 0 $ in measure. –  newbie Jun 19 '11 at 15:00
    
@newbie: $\{|X1_{\Lambda_n}| \geq \varepsilon\} \subset \Lambda_n$. –  t.b. Jun 19 '11 at 15:03

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 $P(\Lambda_n)\leq \frac{\varepsilon}{2n_0}$.

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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 $\lim _{n \to \infty } \int_{\Lambda _n } {X{\rm d\mathbb P}} = 0$.)

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I have five words for you: absolute continuity of the integral. :)

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