Consider the space $\mathcal M$ of all finite complex Borel measures on a segment with norm $\|\mu\|=\int d\,|\mu|$. Assume that a norm-closed linear subspace $\mathcal M_0$ of $\mathcal M$ has the following property:

if $\mu\in\mathcal M_0$ and $f\in L^1(|\mu|)$, then $f\mu\in \mathcal M_0$.

Can $\mathcal M_0$ be characterized as the family of all measures that vanish on a certain fixed collection of subsets of the segment?

Example: the class of all measures $\mu$ such that $d\mu=f\,dx$ for some $f\in L^1$ coincides with the class of measures that vanish on all subsets of zero Lebesgue measure.

Update: For any fixed measure $\mu$, one can define $\mathcal M_0$ as the class of all measures that are absolutely continuous wrt $\mu$. Then the class of subsets in question is the class of subsets of zero $\mu$-measure.

  • $\begingroup$ Fix the interval to be $(0,1)$, is the standard Lebesgue measure in the closed linear span of $\delta_x$ where $x$ ranges over all of $(0,1)$? $\endgroup$ – Willie Wong Feb 19 '15 at 17:10
  • $\begingroup$ I have changed the order of words, that's what I meant -ok? $\endgroup$ – limanac Feb 19 '15 at 17:18
  • $\begingroup$ Nope. The closed linear span of all Dirac measures coincides with the set of all discrete measures. $\endgroup$ – limanac Feb 19 '15 at 17:20
  • $\begingroup$ @Willie Wong: nothing guarantees this; as far as In know, this is not true, and that is why I think that the answer should be negative. $\endgroup$ – limanac Feb 19 '15 at 18:07
  • $\begingroup$ Sorry for the wrong answer (which I deleted). Another try: If $\mathcal M_0$ is the space of all discrete measures then $\emptyset$ is the the only set so that each $\mu \in \mathcal M_0$ vanishes on it. Am I missing something? $\endgroup$ – Jochen Feb 20 '15 at 7:50

Let $\mathcal M_0$ be the space of all discrete measures $\mu$ (i.e. $|\mu|(A)=0$ for some $A$ with countable complement). This is a norm-closed subspace satisfying the property. But $\emptyset$ is the only set on which every discrete measure vanishes.

  • $\begingroup$ @limanac The comment is no longer relevan and should be deleted. $\endgroup$ – Jochen Feb 20 '15 at 8:18
  • $\begingroup$ That was because you undeleted your old answer and my comment appeared automatically. Done. $\endgroup$ – limanac Feb 20 '15 at 8:41

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.