Let $\{r_n \}_{n=1}^\infty$ be an enumeration of the rationals $\mathbb{Q}$. Then:
\begin{align*}
f(x) &= \begin{cases}
x^{-1/2} & \text{if $0 < x < 1$} \\
0 & \text{otherwise} \\
\end{cases} \\
F(x) &= \sum\limits_{n=1}^\infty 2^{-n} f(x – r_n) \\
\end{align*}
Prove that $F$ is integrable and converges for almost every $x \in \mathbb{R}$. Prove that $F$ is unbounded on every interval.
I can prove that $F$ is unbounded on any interval. I am having trouble showing that this converges for every $x$ and is integrable. I found this same question posted previously, but there is no answer and the one hint doesn't seem to apply. $F(x)$ and the terms of its series don't seem to be monotonic.
$F(x)=\sum_{n=1}^{\infty}2^{-n}f(x-r_n)$ is integrable
To show that that $F$ is unbounded on any interval, consider any bound $M$, and any interval $I$ which contains at least one rational $r_n$. Then there is some $x \in I, x \in (r_n, r_n + 1/(2^{2n} \cdot M^2))$ such that:
\begin{align*}
0 < x – r_n &< 1/(2^{2n} \cdot M^2) < 1 \\
2^{-n} (x – r_n)^{-1/2} &> M \\
F(x) &> M \\
\end{align*}
Best Answer
First, we note that $f$ is non-negative and certainly belongs to $L^1(\mathbb{R})$. Given $N \geq 1$, we define $$ F_N(x) := \sum_{n=1}^N 2^{-n}f(x-r_n) $$ which is also non-negative and integrable. Clearly, $F_N(x) \to F(x)$ for almost every $x$. Thus, by Fatou's lemma, we find that \begin{align*} \int_\mathbb{R} F(x)\,\mathrm{d}m &\leq \liminf_{N \to \infty}\int_\mathbb{R} F_N(x)\,\mathrm{d}m\\ &= \liminf_{N \to \infty} \sum_{n=1}^N \int_{\mathbb{R}} 2^{-n}f(x-r_n)\,\mathrm{d}m\\ &=\liminf_{N \to \infty} \sum_{n=1}^N 2^{-n} \left( \int_{\mathbb{R}} f\,\mathrm{d}m\right)\\ &= \left\Vert f\right\Vert_{L^1(\mathbb{R})} \sum_{n=1}^\infty 2^{-n} < \infty. \end{align*} Note we have made use of the translation invariance of the Lebesgue integral. Hence, $F$ is integrable and must therefore be finite almost everywhere.