Problem: Suppose $f \in L^1(\mathbb{R})$. Show that the series $$f(x) = \sum_{n=1}^{\infty}\frac{f(x - \sqrt{n})}{\sqrt{n}}$$ converges for almost all $x \in \mathbb{R}$.
Attempt: Consider the function $F \colon \mathbb{R} \to [0,\infty]$ given by $$F(x) = \sum_{n=1}^{\infty}\frac{\vert f(x - \sqrt{n})\vert}{\sqrt{n}}.$$ We want to show that $F(x) < \infty$ almost everywhere. We show that $F$ is integrable on $[0,1]$. By translating $f$, we may then assume that $F$ is integrable on all intervals of length $1$, from which the desired result follows.
By the monotone convergence theorem, we have $$\int_0^1F(x) \, dx = \sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\int_0^1\vert f(x - \sqrt{n}) \vert \, dx = \sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\int_{I_n} \vert f(x) \vert \, dx,$$ where $I_n = (-\sqrt{n},1-\sqrt{n})$. We can partition $\bigcup_{n=1}^{\infty}I_n$ into disjoint intervals $J_n$ each of which intersects $I_k$ only for $k$ in a subset $S_n$ of $S_m = \{m^2,m^2+1,\ldots,(m+1)^2 - 1\}$ for some $m$. Then we have $$\int_0^1 F(x) \, dx = \sum_{n=1}^{\infty}\sum_{k \in S_n}\frac{1}{\sqrt{k}}\int_{J_n}\vert f(x) \vert \, dx.$$ Since $\vert S_m \vert = 2m + 1$, for each $n$ we have $$\sum_{k \in S_n}\frac{1}{\sqrt{k}} \leq \sum_{n=m^2}^{(m + 1)^2 - 1}\frac{1}{\sqrt{n}} \leq \frac{2m + 1}{\sqrt{m^2}} \leq C$$ for some constant $C > 0$ independent of $m$. Since the $J_n$ are disjoint, we conclude that $$\int_0^1F(x) \, dx \leq \sum_{n=1}^{\infty}C\int_{J_n}\vert f(x) \vert \, dx \leq C\Vert f \Vert_1.$$ This completes the proof since $f \in L^1(\mathbb{R})$.