Suppose $(X,\mathcal{A})$ is a measurable space and $\mathcal{B}$ is an arbitrary subset of $\mathcal{A}$. There exists two finite measures $\mu_1$ and $\mu_2$ on $(X,\mathcal{A})$ such that $\mu_1(A)=\mu_2(A)$ for all $A \in \mathcal{B}$. Is it true that $\mu_1(A)=\mu_2(A)$ for all $A \in \sigma(\mathcal{B})$? And is it true when $\mu_1$ and $\mu_2$ are $\sigma$-finite?

Can anyone give me some hints for this problem I will be really appreciated.


I think you're missing some information or the statement is false. For, consider $X=[0,1]$ and let $\mathcal{A}$ to be the Borel $\sigma$-algebra of subsets of $[0,1]$. Let $\mu_1$ be the Lebesgue measure on $\mathcal{A}$ and $\mu_2=(1/2)\mu_1$. Let $\mathcal{B}$ be the subsets of measure zero.

Note that we obviously have that $\mu_1(A)=\mu_2(A)$ for every $A\in\mathcal{B}$ but $\mu_1(X)\neq \mu_2(X)$ even though $X\in\sigma(\mathcal{B})$.

I haven't found an specific counterexample to case when the measures are $\sigma$-finite, but I think is false because we can't be sure that they are equal on the sets that are described in the definition of $\sigma$-finite measure.

  • $\begingroup$ For the 5th line I think you are meaning $X \in \sigma(\mathcal{B})$. And thank you for your answer! $\endgroup$ – Nullhoz Jan 15 '14 at 21:26
  • $\begingroup$ Yes, you're right. I corrected it. You're welcome $\endgroup$ – Brandon Jan 15 '14 at 21:45

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