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I am working on the following question. I believe I am almost done, but I still have a hole in my solution:

Let $\{r_k\}_{k=1}^{\infty}$, a counting of $\mathbb{Q}$. For every $k \in \mathbb{N}$, let: $$ f_k(x):= \begin{cases} (x - r_k)^{-1/2} &\text{ if } r_k < x \leq r_k + 1 \\ 0 &\text{ else } \end{cases} $$

Prove that $\sum_{k = 1}^{\infty} 2^{-k}f_k$ converges almost everywhere on $\mathbb{R}$ to an integrable function, $f$, such that for every interval $(a, b)$, and every M, $$ m((a,b) \cap \{x:f(x) \geq M\}) > 0. $$

My question is why does $\sum_{k = 1}^{\infty} 2^{-k}f_k$ converges to $f$ almost everywhere and not simply pointwise?

This is what I have so far:

  • $\frac{1}{\sqrt{x}}$ is integrable, hence $f$ is integrable (because its integral on $(0,1)$ equals 2, and $f$'s integral on $\mathbb{R}$ is $2 \times \sum_{k = 1}^{\infty} 2^{-k} < \infty$ by monotone convergence).
  • For every $(a, b)$ there's a $k$ such that $r_k \in (a,b)$, and there is a $\delta$ such that for every $x \in (0, \delta)$, $\frac{1}{\sqrt{x}} > M\times 2^{-k}$ and from this follows the second claim.
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pointwise convergence fails due to unboundedness of summands for fixed $x$ –  no identity Aug 5 '12 at 18:32
    
@Norbert: Any simple argument which would guarantee that there exists some $x$ at which the summands are unbounded? –  Did Aug 5 '12 at 18:45
    
@did, see my answer to this question –  no identity Aug 5 '12 at 19:22
    
Sure, this particular enumeration yields one point with pointwise divergence. But your previous comment says much more, namely that such a point exists for any enumeration. –  Did Aug 5 '12 at 19:27
    
I' haven't said this. –  no identity Aug 5 '12 at 19:30

1 Answer 1

up vote 2 down vote accepted

Let $\{s_m:m\in\mathbb{N}\}$ be enumeration of $\mathbb{Q}\setminus\{-32^{-m}:m\in\mathbb{N}\}$. Define enumeration of $\mathbb{Q}$: $$ r_k= \begin{cases} -32^{-m}&\quad\text{ if }\quad k=2m\\ s_m&\quad\text{ if }\quad k=2m-1 \end{cases} $$ Then $$ f(0)=\sum\limits_{k=1}^\infty 2^{-k}f_k(0)\geq \sum\limits_{m=1}^\infty 2^{-2m}f_{2m}(0)= \sum\limits_{m=1}^\infty 2^{-2m}(0-r_{2m})^{-1/2}\boldsymbol{1}_{[r_{2m},r_{2m}+1)}(0)= \sum\limits_{m=1}^\infty 2^{-2m}(32^{-m})^{-1/2}\boldsymbol{1}_{[-32^{-m},-32^{-m}+1)}(0)= \sum\limits_{m=1}^\infty 2^{-2m}(32^{-m})^{-1/2}= \sum\limits_{m=1}^\infty 2^{m/2}=+\infty $$ Thus you see that for a particular enumeration of $\mathbb{Q}$ there are points $x$ (for our case $x=0$) for which the series $$ \sum\limits_{k=1}^\infty 2^{-k}f_k(x) $$ diverges.

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