Problem: suppose $f\in L_1([0, \infty), \lambda)$, with $\lambda$ the Lebesgue measure. Show that for $\lambda$-almost all $t\in [0, \infty)$ the series $\sum_{n=1}^{\infty}f(nt)$ converges.
This is straightforward if $f$ is continuous, or alternatively if we can show that $f(x)\in L_1$ implies $f(t\cdot \lfloor x/t\rfloor)\in L_1$. Using simple functions here and the Monotone Convergence Theorem also seem like a good idea, except that it's hard to bound $f$ by a simple function whose integral yields a bound on $\sum_{n=1}^{\infty}f(nt)$. Any ideas?
Let $F : [0, \infty) \to [0, \infty]$ be defined by
$$ F(t) = \sum_{n=1}^{\infty} |f(nt)|. $$
It suffices to show that $F(t) < \infty$ for a.e. $t \in [0, \infty)$. To this end, we will show that
Once this claim is established, we can appeal to a standard argument as follows:
For $n = 2, 3, \cdots$, the inequality $\int_{1/n}^{n} F(t) \, dt < \infty$ implies that $F(t)$ is finite $\lambda$-a.e. on $[\frac{1}{n}, n]$. Thus for each $n$, we can pick an exceptional set $E_n \subseteq [\frac{1}{n}, n]$ with $\lambda(E_n) = 0$ such that $F(t) < \infty$ on $[\frac{1}{n}, n] \setminus E_n$. Let $E = \cup_n E_n$. Then it is straightforward to check that $\lambda(E) = 0$ and $F(t) < \infty$ for all $ t \in (0, \infty) \setminus E$.
So it remains to prove the claim. Let $0 < a < b < \infty$ and write
$$ \int_{a}^{b} F(t) \, dt = \int_{a}^{b} \sum_{n=1}^{\infty} |f(nt)| \, dt \stackrel{(*)}{=} \sum_{n=1}^{\infty} \int_{a}^{b} |f(nt)| \, dt. $$
The identity $\text{(*)}$ follows from the monotone convergence theorem (and thus it holds regardless of whether they are finite or not). Now apply the substitution $nt \mapsto t$ to get
\begin{align*} \int_{a}^{b} F(t) \, dt &= \sum_{n=1}^{\infty} \frac{1}{n} \int_{na}^{nb} |f(t)| \, dt \\ &= \sum_{n=1}^{\infty} \frac{1}{n} \int_{(0,\infty)} |f(t)| \mathbf{1}_{[na,nb]}(t) \, dt \\ &= \int_{(0,\infty)} |f(t)| \bigg( \sum_{n=1}^{\infty} \frac{1}{n} \mathbf{1}_{[na,nb]}(t) \bigg) \, dt. \end{align*}
Here, we utilized the monotone convergence theorem again to establish the last equality. Now we show that the integrand is bounded above by a constant multiple of $|f(t)|$. To this end, notice that $na \leq t \leq nb$ is equivalent to $\frac{t}{b} \leq n \leq \frac{t}{a}$. Thus
$$ \sum_{n=1}^{\infty} \frac{1}{n} \mathbf{1}_{[na,nb]}(t) = \sum_{\substack{t/b \leq n \leq t/a \\ n \geq 1}} \frac{1}{n}. $$
Using the estimate $\log (n+1) \leq \sum_{k=1}^{n} \frac{1}{k} \leq 1 + \log n$, it is easy to check that the sum above is bounded by $1 + \log(b/a)$. Therefore
$$ \int_{a}^{b} F(t) \, dt \leq \big( 1+\log\big( \tfrac{b}{a} \big) \big) \int_{(0,\infty)} |f(t)| \, dt < \infty $$
and the claim follows.