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.

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.

share|improve this question
2  
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

2 Answers 2

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.

share|improve this answer

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 $X$.

share|improve this answer
    
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

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.