# Show that $f_n$ is pointwise convergent a.e. using Borell-Cantelli Lemma.

Let $$f_n$$ be a sequence of real valued measurable functions such that for each $$n\in \Bbb N$$

$$\mu\{x\in X:|f_n(x)-f_{n+1}(x)|>\frac{1}{2^n}\}=\frac{1}{2^n}$$.

Show that $$f_n$$ is pointwise convergent a.e. using Borell-Cantelli Lemma.

I dont understand how to use Borel Cantelli Lemma which states that if $$\sum \mu(A_n)<\infty$$ then $$\mu (\lim_{n\to \infty} \sup A_n)=0$$

Can I get some hints ?

• Another way to describe the conclusion about the $\limsup$ of a set is $$\mu\left(x \in X: \{|f_n(x)-f_{n+1}(x)|>1/2^n\} \mbox{ for infinitely many indices n}\right)=0$$ That is $$\limsup_{n\rightarrow\infty} A_n = \{x \in X : \mbox{x is in infinitely many sets A_n}\}$$ – Michael May 1 '19 at 5:50

Borel Cantelli Lemma tells you that with probability $$1$$, $$|f_{n+1}(x)-f_n(x)| \leq \frac 1 {2^{n+1}}$$ for $$n$$ sufficiently large Now use the following:
if $$(a_n)$$ is a sequence of real numbers such that $$\sum |a_{n+1}-a_n| <\infty$$ then the sequence is convergent.
[ $$|a_n-a_{n+m}| \leq |a_n-a_{n+1}|+|a_{n+1}-a_{n+2}|+...+|a_{n+m-1}-a_{n+m}|$$. Can you show that this quantity tends to $$0$$ as $$n,m \to \infty$$?
More details: let $$A_n=\{x:|f_{n+1}(x)-f_n(x)| >\frac 1 {2^{n}}\}$$. Then $$\sum \mu(A_n) <\infty$$. By Borel Cantelli Lemma $$\mu (\lim \sup A_n)=0$$. Let $$E=\lim \sup A_n$$. Then $$\mu (E)=0$$. Suppose $$x \notin E$$. Using the definition of limsup observe that above basic real analysis lemma can be applied to the sequence $$a_n=f_n(x)$$. Hence $$\lim f_n(x)$$ exists whenever $$x \notin E$$.
• Yes, $\sum \frac 1 {2^{n+1}}$ is convergent so its tails sums tend to $0$. – Kavi Rama Murthy May 1 '19 at 7:52