Suppose $\mu$ and $\nu$ are finite positive measures on a measurable space $(X,\mathcal A)$. Let $\mathcal G$ be an algebra of $\mathcal A$. If $\mu$ and $\nu$ are equivalent on $\mathcal G$ in the sense that $ \mu(A)=0$ if and only if $\nu(A)=0$ for all $A\in \mathcal G$, can we conclude that they are equivalent on $\sigma(\mathcal G)$ as well, where $\sigma(\mathcal G)$ is the $\sigma$-algebra generated by $\mathcal G$? If not, is there any counter example?


Thanks for your help!



Equivalent measures on algebra also equivalent on $\sigma$-algebra?
No. Let $X$ be the unit interval $(0,1]$ with its Borel $\sigma$-algebra $\mathcal{A}$. Let $\mu$ be Lebesgue measure. Enumerate the rationals in $(0,1]$ as $\{q\_n\}$ and let $\nu$ be a measure assigning mass $2^{-n}$ to the point $q\_n$. Let $\mathcal{G}$ be the algebra of all finite union of half-open intervals $(a,b]$, which generates $\mathcal{A}$. Then $\mu$ and $\nu$ both assign nonzero measure to every set in $\mathcal{G}$, but are mutually singular on $\mathcal{A}$.


(I was already typing this before Anthony Quas commented!)


