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 ?
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 $x \notin E$.