4
$\begingroup$

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.

  • 0
    Hi im trying this problem now and I was wondering if you could give me a hint on how you showed that if $\sum _{n=1}^{\infty }a_{n}$ converges absolutely then $f$ is integrable2016-06-05

1 Answers 1

2

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.

  • 0
    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 ?2012-09-15
  • 0
    If you knew it, why not include it in the question?2012-09-15
  • 0
    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)2012-09-15
  • 0
    @contd to disprove this statement do n't we need to show "A can hold even if B is false" ?2012-09-15
  • 0
    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). ../..2012-09-15
  • 0
    ../.. Having said that, this is a free market so, if you do not like them, do not buy them... :-) // In the case at hand, I repeat that you ought to have included in your question the example you knew. Finally, as regards the direct sense of the implication (if $f$ is integrable, then...), my answer does not address it and I fail to see this as a problem. Just say so if it is.2012-09-15
  • 0
    I agree with your first point but not entirely with the second, look you are clearly a very knowledgeable mathematician. I want to encourage you to post answers to my questions as that creates the opportunity for me to learn from you. I wish to specialize in analysis, Probability and stochastic process and based on your profile, you know a few things about that stuff, so i suspect many future thread discussions with u :-) but in all sincerity your replies do not provide any motivational reasoning, why to consider something and it's connection with the original question (Contd)2012-09-15
  • 0
    this leaves me guessing, which i find is counter productive. This maybe because i am not mathematically mature like you, but it would make an enormous difference in shaping my mind if you helped highlight the connections. See i know of the alternating series but had not connected it with the question. I am looking to find a counter example where f is integrable but the series does not converge. Now of what i recall Any series which converges absolutely converges conditionally but not vice-versa. So based on this i now know that the f should be such that the series generated diverges (contd)2012-09-15
  • 0
    absolutely and conditionally. If you could highlight how to get to what i want to get to from your reply it would make my life simpler and improve the efficiency of my time. I do have other questions to attempt.2012-09-15
  • 0
    You disagree with *it is counterproductive to leave nothing to do to the learner*? Really?? Wow... I guess we will have to agree that we disagree on this point. // Anyway, are there still some parts of your question whose answer is not clear in your mind? Your last comments leave me the impression that the answer is "No" but I might be wrong.2012-09-15
  • 0
    Don't we want examples where f is integer able and the series diverges instead of where f is not integrable and the series converges ? I think your examples fit the later case which is not what I was asking, I can n't seem bridge this gap please give me an example of the first case. With that said I found it interesting that although the first example is not lebesgue integrable the limit of n going to infinity of the definite integral with upper limit n exists. I wonder if that has any implication about the continuity of integration.2012-09-15
  • 0
    This is impossible since, if $f$ is Lebesgue integrable then $\sum\limits_na_n$ converges absolutely. Hint: $|a_n|\leqslant\int\limits_n^{n+1}|f|$.2012-09-15
  • 0
    Great. $ $ $ $ $ $2012-09-15
  • 0
    @Did, $f(x)=(-1)^{n}/n$ is not a function of $x$.2015-11-19
  • 0
    @JessyCat What? Please read carefully the post (in particular, stopping at mid-sentence can be disastrous...).2015-11-19