The problem (not homework) I practiced is
Consider the probability space $([0,1], B_{[0,1]}, P)$ where $B_{[0,1]}$ is the Borel set and $P$ is Lebesgue measure on $[0,1]$. For any integer $n>0$, there exist $m$ and $k$ such that $n = 2^m-2+k$ and $0\leq k \leq 2^{m+1.}$ Define
$$X_n(\omega)=\left\{\begin{array}{clcr} 1, \mbox{} \frac{k}{2^m}\leq \omega\leq\frac{k+1}{2^m}\\0, \mbox{ otherwise} \end{array}\right.$$ for any $\omega\in[0,1]$. Is the statement, $X_n \to 0$ a.s., correct?
I know the solution can be found from p.36, Exercise 47, Mathematical Statistics: Exercises and Solutions by Jun Shao. That is, $X_n(\omega)$ does not converge to $0$ a.s. The reason is following.
For any fixed $\omega \in[0,1]$ and m, there exists $k$ with $1\leq k\leq 2^m$ such $\frac{k-1}{2^m}\leq \omega \leq \frac{k}{2^m}$. Let $n_m =2^m-2+k$. Then $X_{n_m}(\omega)=1$. Since m is arbitrary selected we can find an infinite sequence $\{n_m\}$ such that $X_{n_m}(\omega)=1$. This implies $X_n{\omega}$ does not converge to 0. Since $\omega$ is arbitrary, $X_n$ does not converge to 0 a.s.
However, I got the different answer. My answer is :
Since $P(|X_n| > \epsilon) \leq \frac{1}{2^m}$ for any $\epsilon >0$, $\sum_n P(|X_n| > \epsilon) < \infty$. By the first Borel-Cantelli lemma, $P(\{|X_n| > \epsilon\} \mbox{ i.o.}) = 0$. Since $\epsilon$ is arbitrary, $X_n \to 0$ a.s.
If my answer is wrong, I'm curious why the 1st Borel Cantelli lemma does not work or where something wrong is.
