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Consider $f$ a bounded, measurable function defined on $[1,\infty]$ and define $a_n := \int_{[n,n+1)} f$. Does the fact that $f$ is integrable imply $\sum_{n=1}^\infty a_n$ converges?

There has been a post regarding this question but I would like to argue that the argument \begin{equation} \sum^\infty_{n=1} a_n = \int^\infty_1 f \end{equation} made in that post is not correct.

My attempt was the following: Since $f$ is integrable, $|f|$ is also integrable. Therefore, \begin{equation} \int_{[1,\infty)} |f| = \lim_{k\to \infty} \int^k_1 |f| < \infty. \end{equation} Moreover, we have the following: \begin{align} \sum^\infty_{n=1} a_n &= \sum^\infty_{n=1} \int^{n+1}_n f \\ &\leq \sum^\infty_{n=1} |\int^{n+1}_n f| \\ &\leq \sum^\infty_{n=1} \int^{n+1}_n |f| \\ &=\lim_{k\to\infty} \sum^k_{n=1} \int^{n+1}_n |f| \\ \end{align} However, I cannot proceed further from here. Can anyone help?

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  • $\begingroup$ "Integrable" means $\int |f|<\infty$, right? $\endgroup$
    – paul garrett
    Commented Mar 21, 2022 at 0:42
  • $\begingroup$ @paulgarrett I am talking about Lebesgue integration. As far as I know, yes. A function $f$ is integrable means that $\int f < \infty$. Also, a function $f$ is integrable if and only if $|f|$ is integrable. $\endgroup$
    – qqq123
    Commented Mar 21, 2022 at 0:55
  • $\begingroup$ Ok, then, yes, $\int |f|<\infty$ does imply that $\int f$ exists and is finite. But/and then I don't understand why you'd think that breaking the integral of $f$ into pieces is not ok. Yes, if we make no assumptions on (absolute!) integrability of $f$, then the integrals over intervals can have self-cancellations, like $(1-1)+(1-1)+...=0$. But, depending on what you mean by "integrable", this would be assumed-away... Can you clarify? $\endgroup$
    – paul garrett
    Commented Mar 21, 2022 at 1:01
  • $\begingroup$ From your comments below, perhaps there is a minor issue of "logic". Namely, yes, $\sum_n \int_n^{n+1} f$ convergent does not imply that $\int_1^\infty f$ is convergent, as your example of $\sin 2\pi x$ (or similar) illustrates. But $\sin 2\pi x$ is neither (Lebesgue) integrable, nor is it "absolutely" integrable, so it is not surprising that there's this seeming discrepancy. In the end, I'd recommend caution about use of "integrable", without "absolutely", since many Lebesgue-theory introductions allow an "integrable" non-negative-real-valued function to have integral $+\infty$... $\endgroup$
    – paul garrett
    Commented Mar 21, 2022 at 1:06
  • $\begingroup$ ... AND I'd claim that there's no genuine subtlety or pathology here, just an easily-understood technicality that is mostly not the salient issue in real-life situations. :) $\endgroup$
    – paul garrett
    Commented Mar 21, 2022 at 1:08

3 Answers 3

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We have \begin{align} \sum_{n = 1}^{\infty}|a_n| &\leq \sum_{n = 1}^{\infty}\int_{[n, n + 1)}|f| \\ &= \sum_{n = 1}^{\infty}\int |f| 1_{[n, n + 1)} \\ &= \int \sum_{n = 1}^{\infty}|f| 1_{[n, n + 1)} \\ &= \int |f|1_{[1, \infty)} \\ &< \infty. \end{align} The second equality is by MCT and additivity of the integral on nonnegative functions. Alternatively, you can argue directly using DCT that $\sum_{n = 1}^{\infty}a_n = \int f1_{[1, \infty)}$.

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Since $f$ is integrable, you know that $$ \int\limits_1^{k+1}|f|\,dx=\sum_{n=1}^{k}\int\limits_n^{n+1}|f|\,dx $$ converges to $\int_1^\infty |f|\,dx<\infty$ as $k\to\infty$ (by monotonic convergence, extending $f$ by zero beyond $k+1$). Thus your last limit is finite and your series $\sum a_n$ is convergent

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  • $\begingroup$ Isn't this statement equivalent to $\sum^\infty_{n=1} a_n = \int^\infty_1 f$? But $\sum^\infty_{n=1} a_n = \int^\infty_1 f$ is not correct. A counterexample is set $f:= \sin(2\pi x)$. $a_n$ is zero for every $n$ and therefore $\sum a_n$ is zero. But $\sin(2\pi x)$ is not integrable. Assume that this statement is correct, then the fact that $\sum a_n$ converges would imply $\int f$ is integrable, which contradicts to the example I made. $\endgroup$
    – qqq123
    Commented Mar 20, 2022 at 21:15
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    $\begingroup$ No, it is not. What we are saying is that $\sum |a_n|$ is convergent, and its sum is bounded by $\int|f|\, dx$. Therefore $\sum a_n$ is also convergent. $\endgroup$
    – GReyes
    Commented Mar 20, 2022 at 21:39
  • $\begingroup$ I get the fact that $\sum |a_n| $ is convergent, but this is what we are trying to prove. I didn't really get how you approach the statement in your answer by monotonic convergence. I assume that you meant Monotone Convergence Theorem. But this theorem requires that the sequence to be bounded. However, we are trying to prove boundedness. $\endgroup$
    – qqq123
    Commented Mar 20, 2022 at 22:12
  • $\begingroup$ Absolute convergence implies convergence. Yes, I mean Monotone convergence theorem. The theorem, for non-negative functions like $|f|$, says that the limit of the integrals is the integral of the limit. $|f|$ is the point-wise monotonic limit of its restrictions to $[1,n+1)$. Your limit is integrable, so it it finite. $\endgroup$
    – GReyes
    Commented Mar 20, 2022 at 22:14
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    $\begingroup$ Observe that $\sum_{n=1}^k\int_{n}^{n+1}|f|dx=\int_1^{k+1}|f|dx$. We apply MCT to the sequence $g_k(x)=|f|\chi_{[1,k+1)}(x)$. Clearly, $g_k$ approaches monotonically to $|f|$. Therefore, $\int g_k(x)\,dx\to\int|f|\,dx<\infty$ as $k\to\infty$. $\endgroup$
    – GReyes
    Commented Mar 21, 2022 at 0:12
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I do think that a substantial part of the issue here is terminological, or similar. But not a mysterious mathematical issue (beyond a standard sort of example).

Namely, first, indeed, convergence, or even absolute convergence, of the series $\sum_n \int_n^{n+1} f(x)\;dx$ proves nothing about any sort of convergence or integrability or absolute integrability of $f$. The example of $f(x)=\sin 2\pi x$ illustrates this, and it's not a pathology, etc.

On another hand, equally robustly, if "$f$ integrable" means absolutely integrable, then by various means $\int_1^\infty f = \sum_n \int_n^{n+1} f$.

If real-valued $f$ is "integrable" in a technical Lebesgue-theory sense, namely, that one or the other of its positive and negative parts is absolutely integrable, while the other may fail to be so... we still reach the conclusion.

If there are other contexts of interest to the questioner, I'd be happy to entertain them. But, beyond the basic counter-example to a too-naive appraisal of the situation, I do not think there's a pathology here.

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