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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.

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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
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.

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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

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