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How would I go about finding a family $X$ of Borel sets in $\mathbb{R}$ that generate the Borel $\sigma$-algbera on $\mathbb{R}$ and two finite Borel measure $\mu$ and $\nu$ that agree on $X$ but do not agree on the whole Borel $\sigma$-algebra.

I know that $X$ cannot be a $\Pi$-system, so I was thinking of using the open intervals but I'm really struggling with the measures. The only finite measures I can think of are Dirac point measures.

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Take $X$ to be the open intervals not containing $0$, one measure the zero measure and the other the Dirac point measure at $0$. – t.b. Nov 3 '11 at 16:41
I don't know if it may be helpful, but getting a finite measure is not so hard - in one way, you can use $\arctan$ to contract $\mathbb{R}$ to $(-\pi/2, \pi/2)$ and measure contracted sets or you can use integrals to define the measure of a set. Any integrable and non-negative function is OK. – savick01 Nov 3 '11 at 16:49
Now t.b. has an example where $X$ is a $\Pi$-system, but $\mu(\mathbb R) \ne \nu(\mathbb R)$. How about another example where $X$ is not a $\Pi$-system and $\mu(\mathbb R) = \nu(\mathbb R)$ ?? – GEdgar Jul 25 '13 at 12:08

Is this question really 8 months old? Try this:

Let $\mathcal X$ consist of all borel sets $A \subseteq \mathbb R$ such that $$ \text{either}\qquad A \cap \{1,2,3,4\} = \{1,2\}\qquad\text{or}\qquad A \cap \{1,2,3,4\} = \{1,3\} . $$ Let $\mu = \delta_1+\delta_4$. That is, points $1$ and $4$ each have measure $1$, everything else measure zero. And let $\nu = \delta_2 + \delta_3$.

Show: (a) $\mu(A)=\nu(A)$ for all $A \in \mathcal X$. (b) $\sigma(\mathcal X)$ is all Borel sets.

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See t.b.'s another construction: Take $X$ to be the open intervals of $R$ not containing $0$ and $1$, one measure Dirac point measure at $0$ and the other the Dirac point measure at $1$. Both measures are nontrivial and agree on $X$.

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A tricky one! This $X$ does not generate the Borel sigma-algebra. Any set in it either contains both $0,1$ or neither. – GEdgar Jul 25 '13 at 12:01

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