1
$\begingroup$

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!

$\endgroup$
1
  • 1
    $\begingroup$ No. Generating algebras can be quite small. e.g. finite unions of $[a,b)$ in $[0,1)$. It's quite easy to come up with two measures that give positive measure to all open sets. $\endgroup$ Oct 19, 2015 at 2:21

1 Answer 1

4
$\begingroup$

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!)

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.