Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\mu_1$ and $\mu_2$ finite measures on $\sigma$-algebra $\mathfrak B$ such that $\mu_1(X)=\mu_2(X)$, and $\mathcal A$ an intersection stable generator of $\mathfrak B$ such that $\mu_1(A)=\mu_2(A)$ for all $A\in\mathcal A$. It is well known that above hypothesis implies $\mu_1=\mu_2$. Can we have the same conclusion if $\mu_1$ and $\mu_2$ are totally finite signed measures? Thanks.

share|improve this question
    
Do you still assume that $\mu_1(X)=\mu_2(X)$? In this case why wouldn't the same argument work? –  Davide Giraudo Aug 12 '12 at 17:09
    
I drop that assumption. –  Deco Aug 12 '12 at 19:28
add comment

1 Answer

up vote 3 down vote accepted

Take $X$ an uncountable set, $\mathcal B$ the $\sigma$-algebra of countable sets and their complement, $\mathcal A$ the collection of countable subsets of $X$. Take $$\mu_1(A):=\begin{cases}0&\mbox{ if }A\mbox{ is countable},\\ 1&\mbox{ if }X\setminus A\mbox{ is countable,} \end{cases}$$ and $\mu_2:=2\mu_1$. $\mu_1$ and $\mu_2$ are finite measures, and $\mu_2-\mu_1$ coincides with the $0$ measure on $\mathcal A$ but not on $\mathcal B$.

The main problem is that the measure don't have the same total mass. If we take $\mu_1(X)=\mu_2(X)\in\Bbb R$, then we have $\mu_1=\mu_2$ by a similar argument than in the case of a finite non-negative measure.

share|improve this answer
    
I found this theorem in more than one papers: Let $\mu_1$ and $\mu_2$ be totally finite signed measures on the Borel $\sigma$-algebra $\mathfrak B$ of $\mathbb R^n$, and let $\mathcal A$ an intersection stable generator of $\mathfrak B$ such that $\mu_1(A)=\mu_2(A)$ for each set $A\in\mathcal A$. Then $\mu_1=\mu_2$. –  Deco Aug 13 '12 at 15:13
    
If the generator contains $X$, no problem. But when it's not assumed I think the counter-example works. Do they give a reference in the paper? –  Davide Giraudo Aug 13 '12 at 15:18
    
All of the books i read demand $\mu_1(X)=\mu_2(X)$ for this theorem, it is needed to construct d-system for proof. This make me confuse. –  Deco Aug 13 '12 at 15:21
    
If H1 is the hypothesis: $\mu_1(A)=\mu_2(A)$ for $A$ in an intersection stable generating collection and H2 the hypothesis $\mu_1(X)=\mu_2(X)$, then (H1+H2) gives uniqueness, but not H1 alone. –  Davide Giraudo Aug 13 '12 at 15:22
    
There is no information about $X$ being element of $\mathcal A$. They refer the proof to Maß- und Integrationstheorie. 3., erweiterte Auflage by Jurgen Elstrodt and Measure and integral by Konrad Jacobs. Unfortunately, i can access none of them. –  Deco Aug 13 '12 at 15:27
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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