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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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    I drop that assumption.2012-08-12

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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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    Maybe the fact $\mathfrak B$ being Borel $\sigma$-algebra guarantees $\mu_1(X)=\mu_2(X)$.2012-08-13