# Classes of measures that are closed under multiplication

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.

• 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 $(0,1)$? – Willie Wong Feb 19 '15 at 17:10
• I have changed the order of words, that's what I meant -ok? – limanac Feb 19 '15 at 17:18
• Nope. The closed linear span of all Dirac measures coincides with the set of all discrete measures. – limanac Feb 19 '15 at 17:20
• @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. – limanac Feb 19 '15 at 18:07
• 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? – Jochen Feb 20 '15 at 7:50

## 1 Answer

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.

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