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I want to find an example of two finite measures $\mu$ and $\nu$ on a measure space $(X,S)$ with $\mu(X)=\nu(X)$ such that {$A\in S: \mu(A)=\nu(A)$} is not a $\sigma$-algebra.

Can someone help me?

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up vote 4 down vote accepted

On $X=\{a,b,c,d\}$, choose $\mu$ uniform on $\{a,b\}$ and $\nu$ uniform on $\{c,d\}$. Then $\mu(A)=\nu(A)$ for $A=\{a,c\}$ and $A=\{a,d\}$ but not for $A=\{a\}$. The class $\{A\subseteq X\mid\mu(A)=\nu(A)\}$ is not a sigma-algebra since it is not stable by intersection.

This example can be adapted to every measurable space $(X,S)$ such that $S$ contains four disjoint nonempty subsets of $X$.

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+1 Nice counterexample. – copper.hat Nov 1 '12 at 20:11
@did This is impressive. Thanks! – Frank Lu Nov 1 '12 at 20:29

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