I'm working on the following exercise:

Let $f \in \mathcal L^1([0,\infty), \lambda)$ be a Lebesgue integrable function on $[0,\infty)$. Show that for $\lambda$-almost every $t \in (0,\infty)$, the series $\sum_{n=1}^\infty f(nt)$ converges absolutely.

Here $\lambda$ is Lebesgue measure. I'm not really sure how to go about this. I thought I could use monotone convergence on the functions $F_n = \sum_{k=1}^n f(kt)$, but I have no reason to suspect these are integrable. I'm wondering if there's a Fatou argument I can make instead, but I'm having trouble seeing it.

This exercise comes from Section 4.2 of Achim Klenke's book on probability theory (this section is on Fatou's lemma and monotone convergence). The series in question looks suspiciously similar to a limiting term in a Riemann integral, but this is discussed in the following section, so I'd like to avoid using Riemann integration arguments here. (Though the exercise could just be misplaced.)


1 Answer 1


We will prove that for each $\epsilon>0$ one has $$\int_{\epsilon/2}^{\epsilon} \sum_{n=1}^{\infty} |f(nt)|dt <\infty.\tag{1}$$ This implies your claim easily enough.

To prove (1), let us break up the sum into dyadic partitions, i.e., write $$\sum_{n=1}^{\infty} |f(nt)| = \sum_{k=0}^{\infty} \sum_{j=2^k}^{2^{k+1}-1}|f(jt)|.$$ Setting $g_k(t) := \sum_{j=2^k}^{2^{k+1}-1}|f(jt)|$, one notices that $$\int_{\epsilon/2}^{\epsilon} g_k(t) = \sum_{j=2^k}^{2^{k+1}-1} \int_{\epsilon/2}^{\epsilon}|f(jt)|dt = \sum_{j=2^k}^{2^{k+1}-1} \frac{1}{j} \int_{j\epsilon/2}^{j\epsilon}|f(t)|dt \leq \bigg( \sum_{j=2^k}^{2^{k+1}-1} \frac{1}{j}\bigg)\int_{2^{k-1}\epsilon}^{2^{k+1}\epsilon}|f|\leq \int_{2^{k-1}\epsilon}^{2^{k+1}\epsilon}|f|.$$ In the last bound we used $1/j \le 1/2^k$. Consequently, one finds that $$\int_{\epsilon/2}^{\epsilon} \sum_{k \ge 0}g_k(t)dt \leq \sum_{k \ge 0} \int_{\epsilon 2^{k-1}}^{\epsilon 2^{k+1}}|f(t)|dt \le 2\|f\|_{L^1},$$proving (1).

  • $\begingroup$ I'm not sure I understand the last inequality. Could you please explain? $\endgroup$ Sep 10, 2022 at 15:06
  • $\begingroup$ @Nirai break the interval [2^{-k-1},2^{-k+1}}] into two pieces and collect some repeating terms. $\endgroup$
    – shalop
    Sep 13, 2022 at 3:24

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