# Example that a measurable function $f$ on $[1,\infty )$ can be integrable when $\sum _{n=1}^{\infty }\int_{n}^{n+1}f$ diverges.

I am seeking help in my attempt to formulate a proof to disprove the following.

For a measurable function $f$ on $[1,\infty )$ which is bounded on bounded sets, define $a_n= \int_{n}^{n+1}f$ for each natural number $n$. Is it true that $f$ is integrable over $[1,\infty )$ if and only if the series $\sum _{n=1}^{\infty }a_{n}$ converges ?

I strongly suspect this to be false and could prove that if the series converged absolutely then $f$ is integrable over $[1,\infty )$ but i am unsure how to extend this to answer the question based on conditional convergence of the series.

Any help would be much appreciated.

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## 1 Answer

Try $f(x)=(-1)^n/n$ for every $n\leqslant x\lt n+1$ and $n\geqslant1$. This function $f$ is not Lebesgue integrable. One can modify the example, first, to get $f$ Riemann integrable on compact sets but not Riemann integrable on $[1,+\infty)$, and second (regarding your I could prove paragraph), to get $\sum\limits_n|a_n|$ convergent and $f$ not Lebesgue integrable.

Edit: A second example is $f(x)=1$ if $n\leqslant x\lt n+1/2$ for some $n\geqslant1$ and $f(x)=-1$ otherwise. Then $a_n=0$ for every $n$ hence $\sum\limits_n|a_n|$ converges, $f$ is locally Riemann integrable but not Riemann integrable on $[1,+\infty)$ since $x\mapsto\int\limits_0^xf$ oscillates between $0$ and $1/2$, and $f$ is not Lebesgue integrable either since $|f|=1$ identically.

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thanks i actually knew of the counter example, but i am not sure if i fully understand rest of your answer. Could you please possibly expand a little more ? –  Hardy Sep 15 '12 at 6:01
If you knew it, why not include it in the question? –  Did Sep 15 '12 at 6:07
Might i make a suggestion, if i am not crossing any line here. When you answer elementary questions like the ones i post on this site, Please add reasoning why to do something. The people who post these questions are posting because of lack of that understanding. You might point out the answer but it defeats the purpose. Like now i am struggling to understand how your set of hints describe a solution to my problem. I am new to proofs so please bear with me here. I see the hypothesis as "Is it true, A holds if and only if B" (@contd) –  Hardy Sep 15 '12 at 6:29
@contd to disprove this statement do n't we need to show "A can hold even if B is false" ? –  Hardy Sep 15 '12 at 6:31
Many mathematicians interested in pedagogical matters came to the conclusion that (1.) simple examples disproving natural conjectures are a most efficient way to make learners understand and remember why said conjecture is false, (2.) it is counterproductive to leave nothing to do to the learner. With all due modesty, this is why the present answer is written as it is. In other words, I am QUITE GLAD that now you are struggling to understand how this set of hints describes a solution to your problem (although I would not describe my answer as a set of hints, in the present case). ../.. –  Did Sep 15 '12 at 7:02
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