0
$\begingroup$

Consider a probability measure $d\mu = w(t) dt$ with $w(t)\in L^1(I)$, $I =\left[ 0,1\right]$. What are the minimal assumption I can take on two functions $f,g:I\ \to \mathbb{R}$ so that an integration by parts -like statement could be made, and what would it be?

Also, can the demands on measure space $\left(\Omega, \mu\right)$ be relaxed so that this result would be still valid? Something with non compact $\Omega$, infinite measure etc.

This post asks a somehow related question, but not quite the same question.

Thanks

Amir

Improve this question
$\endgroup$
2
  • $\begingroup$ Perhaps you want the integration by parts for the Lebesgue-Stieltjes integral. en.wikipedia.org/wiki/… $\endgroup$
    – Gerald Edgar
    Commented May 18, 2016 at 23:37
  • $\begingroup$ @GeraldEdgar Isn't there something for functions without bounded variation? $\endgroup$
    – Amir Sagiv
    Commented May 20, 2016 at 6:34

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .