# Almost sure convergence

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.

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.

• What is the reasoning behid your last sentence "Since $\epsilon$ is arbitrary, $X_n \to 0$ a.s"? – user10525 Sep 12 '12 at 10:13
• @Procrastinator, Since $P(\{|X_n| > \epsilon\} i.0) = 0$ implies $P(\{|X_n| \leq \epsilon\} i.0) = 1$ for any $\epsilon >0$, I conclude my last sentence. – Jim Sep 12 '12 at 10:23
• Note that $2^m=n+2-k$, then the convergence of the series is not ensured because the summation is over $n$. Otherwise, you would be correct, see. – user10525 Sep 12 '12 at 11:04
• You left out the important but obvious fact that the limit is taken as n goes to infinity. – Michael R. Chernick Sep 12 '12 at 11:04
• One of the proofs that the harmonic series diverges is that it is at least $1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + 1/16 + ...$ with $2^{n-1}$ copies of $1/2^n$. This series of reciprocals of powers of $2$ diverges in a similar way. – Douglas Zare Sep 12 '12 at 13:59