4
$\begingroup$

The famous Riemann rearrangement theorem states that for a conditionally convergent real number series, we can rearrange the order of summation to make it converge to any prescribed number in the extended real line. In particular, this result sheds much light on the significance of absolute convergence, without which it would be quite dangerous to manipulate a convergent series.

For improper (Riemann) integrals, we can also distinguish conditionally convergent integrals from absolutely convergent ones using analogous definitions. I'm wondering, however, if there also exists an analogous "rearrangement theorem" for improper integrals which reveals the essential difference (like eligibility for rearrangement, in the case of numerical series) between the two kinds of convergent integrals? Indeed, can we even define an integral version of "rearrangement"?

Of course, one distinction I'm already aware of is that absolutely convergent integrals are also integrable in the Lebesgue sense while conditionally convergent ones fail to be. But this is not what I want, since it doesn't appear nearly as striking as what is exhibited in the series version.


PS: as far as I can tell, one probable way to see why eligibility for rearrangement matters is how we define a valid expectation for a numerical random variable. We require expectations (numerical series for discrete random variables, and integrals for continuous ones) to be absolutely convergent, for if they were not, then they would undesirably depend on the "chronological order" in which we observe events, violating our basic principle that expectations should be stable and inherent in the random variable itself, rather than affected by how each event chronologically arises.

$\endgroup$
3
$\begingroup$

Here is a possible definition of a rearrangement: Suppose that we are given a (possibly non-compact) interval $I$. Take a nested sequence of compact sets $I_n$ which limit to $I$. Then consider the sequence of integrals $$\int_{I_n} f(x)dx$$ We may then consider the limit $$\lim_{n\to \infty} \int_{I_n} f(x)dx$$ For an integral such as $\int_{0}^\infty \frac{\sin x}{x}dx$, we might take the "standard" sequence of $I_n$ to be $I_n=[0,n\pi]$. You could, however, let $I_n$ be the union of finite combinations of the odd $[(2n+1)\pi,(2n+2)\pi]$ and even $[2n\pi,(2n+1)\pi]$ intervals, as long as they are nested and eventually cover all of $[0,\infty)$. For a good choice of function and intervals, you should be able to get the standard rearrangement proof to work. (I think that $\sin(x)/x$ and the $I_n$ I mentioned should do the trick.)

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Fair point. Seems that the "rearrangement" of an integral can be done, via partitioning the integration domain, in exactly the same way as we rearrange numerical series. $\endgroup$ – Vim May 13 '16 at 14:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.