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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.

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  • $\begingroup$ What is the reasoning behid your last sentence "Since $\epsilon$ is arbitrary, $X_n \to 0$ a.s"? $\endgroup$ – user10525 Sep 12 '12 at 10:13
  • $\begingroup$ @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. $\endgroup$ – Jim Sep 12 '12 at 10:23
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    $\begingroup$ 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. $\endgroup$ – user10525 Sep 12 '12 at 11:04
  • $\begingroup$ You left out the important but obvious fact that the limit is taken as n goes to infinity. $\endgroup$ – Michael R. Chernick Sep 12 '12 at 11:04
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    $\begingroup$ 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. $\endgroup$ – Douglas Zare Sep 12 '12 at 13:59
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Shao is correct. As Procrastinator points out m has a special form. So you have only shown convergence to 0 along a particular subsequence and not every subsequence.

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