1
$\begingroup$

I'm trying to prove the following:

Show that the integral of an absolutely continuous function, which asymptotically converges to zero (so the function value is zero when its argument is infinity), exists. In other words, I need to show that the integral tends to some constant $c<\infty$.

Any help is greatly appreciated!

Edit It's exactly the opposite of this problem.

$\endgroup$
3
  • $\begingroup$ What's the defintion of an absolutely continuous function on an unbounded interval? $\endgroup$
    – zhw.
    Oct 24, 2016 at 20:10
  • $\begingroup$ I assumed that the definition of absolute continuity on a closed interval can be extended to a (half) open inteval, as discussed in this post. Actually, I'm not a 100% sure if this assumption is justifiable. $\endgroup$
    – rbrns
    Oct 25, 2016 at 7:48
  • $\begingroup$ Maybe assume the (AC) function is of the form $\int _a ^x f(t)\mu(\mathrm{d}t)$? That is, it is a Lebesgue's integral. $\endgroup$
    – Ranc
    Oct 25, 2016 at 7:58

1 Answer 1

Reset to default
1
$\begingroup$

This is clearly not true, as $x \mapsto \frac{1}{x + 1}$ is absolutely continuous over $[0, +\infty[$, tends to zero but has no definite integral.

$\endgroup$
5
  • $\begingroup$ An excellent example, thanks. But are there conditions that the function should satisfy in order for the limit of the integral to exist? $\endgroup$
    – rbrns
    Oct 25, 2016 at 9:31
  • $\begingroup$ well, given that any continuous function with limit 0 at infinity is absolutely continuous, the conditions are the same as for any continuous function with limit 0. $\endgroup$
    – Ulysse Mizrahi
    Oct 25, 2016 at 9:38
  • $\begingroup$ Any idea where I can find those specific conditions? Thanks for your answer. $\endgroup$
    – rbrns
    Oct 25, 2016 at 9:56
  • $\begingroup$ There are no simple conditions unfortunately other than just "be integrable". Also consider accepting the answer if it fits your question. $\endgroup$
    – Ulysse Mizrahi
    Oct 25, 2016 at 11:52
  • $\begingroup$ I understand. Thank you for your effort. $\endgroup$
    – rbrns
    Oct 25, 2016 at 12:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .