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.