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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.
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What do you know? What did you try? – Did Feb 23 '13 at 17:07
What has you stuck? At least one, if not all, of these are straightforward. – Clayton Feb 23 '13 at 17:11
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
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.

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