Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\mu$ be a measure on $X$. We have a sequence of real measurable functions ${f_n}$ on $X$ such that $$\sum_{n=1}^{\infty}\mu (\{x\in X:|f_n(x)|> \tfrac {1}{2^n} \})<\infty.$$ Then what assertion is false:

  1. $\{f_n\}$ a.e. converges to zero.
  2. $\{f_n\}$ does not converge in measure to zero.
  3. $\displaystyle\sum_{n=1}^{\infty}|f_n|$ converges a.e.
share|improve this question
1  
What do you know? What did you try? –  Did Feb 23 '13 at 17:07
1  
What has you stuck? At least one, if not all, of these are straightforward. –  Clayton Feb 23 '13 at 17:11
1  
Hint: Do you know that if $\sum_{n=1}^\infty \mu(E_n) < \infty$ then almost all $x$ lie in at most finitely many of the sets $E_n$? –  Ayman Hourieh Feb 23 '13 at 17:20
    
@AymanHourieh. Yes it is last theorem in second chapter of Rudin's book. –  rese Feb 23 '13 at 17:27
    
@reme Apply the theorem to the sum you have here. What do you conclude? –  Ayman Hourieh Feb 23 '13 at 17:32
show 1 more comment

1 Answer

up vote 1 down vote accepted
  1. You already figured out this one by applying Borel-Cantelli's 0-1-law: Since $\sum_{n=1}^{\infty} \mu(E_n)<\infty$ there exists for almost all $x \in X$ some $N \in \mathbb{N}$ such that for all $n \geq N$ we have $x \notin E_n$, i.e. $$\forall n \geq N: |f_n(x)| \leq \frac{1}{2^n} \tag{1}$$ This implies $f_n \to 0$ $\mu$-almost everywhere.

  2. This one should be obvious from the first part.

  3. Let $x \in X$ and $N \in \mathbb{N}$ such that (1) holds. Then $$\sum_{n=1}^{\infty} |f_n(x)| = \sum_{n=1}^N |f_n(x)| + \sum_{n=N+1}^{\infty} |f_n(x)| \stackrel{(1)}{\leq} \sum_{n=1}^N |f_n(x)| + \sum_{n=N+1}^{\infty} \frac{1}{2^n}$$

    The first series on the right-hand side is convergent since it's a finite sum. The second one is a geometric series, hence in particular convergent. Since (1) holds for almost all $x \in X$ we conclude $\sum_{n=1}^{\infty} |f_n| < \infty$ almost everywhere.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.