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Problem statement: Let $(X,\mathcal{A})$ be a measurable space, and let $\mu$ and $\nu$ be two finite measures. We say $\mu$ and $\nu$ are equivalent measures if $\mu \ll \nu$ and $\nu \ll \mu$ (if $\nu$ and $\mu$ are each absolutely continuous w.r.t. the other). Show that $\mu$ and $\nu$ are equivalent if and only if there exists a $\mu$-integrable function $f$ is that is strictly positive a.e. w.r.t. $\mu$ such that $d \nu = f d\mu$.

My attempt at a solution: For the reverse direction, I tried to use the Radon-Nikodym theorem. Since $\mu$ is $\sigma$-finite, and $\nu$ is finite, and $\nu \ll \mu$, there exists a $\mu$-integrable, non-negative function, measurable function, call it $f$, such that $\nu(A) = \int_A f d\mu$ for all $A \in \mathcal{A}$. Now, we just need to show that this function is strictly positive a.e. to get our conclusion. What I tried was to set $B = \{x : f(x) = 0\}$, and noting that $$\nu(B) = \int_B f d\mu,$$ we find that $\nu(B) = 0$. But $\mu \ll \nu$, so we have that $\mu(B) = 0$, as well, and we are done.

For the forward direction, we assume that there exists a $\mu$-integrable, strictly positive a.e. function w.r.t. $\mu$ such that $d\nu = f d\mu$. Now, this implies that for any measurable set $A$, $$ \nu(A) = \int_Af d\mu.$$ It is clear that $\nu(A) \ll \mu(A)$. Now if $\nu(A) = 0$, we have that $0 = \int_A fd\mu$. What I would like to say there is that this implies that $f = 0$ a.e. on $A$, which I believe I can prove, but that since $f$ is strictly positive a.e., that therefore $\mu(A) = 0$.

Here is my problem: I don't think that I have used that $\mu$ is a finite measure anywhere, which makes me very uncomfortable. I also am a little unsure regarding the argument for the forward direction.

My apologies for posting so many homework-sounding questions in the analysis section lately - I'm studying for a qual in September, so I'm trying to do a bunch of problems before then, and this seems like a great place to get help when I get stuck - hopefully I'm not being too annoying.

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  1. You wrote: "For the forward direction, we assume that there exists a $\mu$-integrable, strictly positive a.e. function w.r.t. $\mu$ such that $d\nu = f d\mu$. Now, this implies that for any measurable set $A$, $$ \nu(A) = \int_Af d\mu.$$ It is clear that $\nu(A) \ll \mu(A)$. Now if $\nu(A) = 0$, we have that $0 = \int_A fd\mu$. What I would like to say there is that this implies that $f = 0$ a.e. on $A$, which I believe I can prove, but that since $f$ is strictly positive a.e., that therefore $\mu(A) = 0$. "

Yes, your argument is OK. You can conclude $f = 0$ a.e. on $A$, because $f$ is strictly positive a.e..

  1. The result is valid even is both $\mu$ and $\nu$ are $\sigma$-finite. The is same. Radon-Nikodym theorem is valid if $\mu$ and $\nu$ are both $\sigma$-finite measures and $\nu \ll \mu$. So, in fact, you don't need neither $\mu$ nor $\nu$ to be finite measures (but you need them to be $\sigma$-finite).
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  • $\begingroup$ Interesting, the book that I am reading gives the theorem with $\mu$ as $\sigma$-finite, and $\nu$ as finite, and definitely supposes both $\mu$ and $\nu$ to be finite in this problem - it's good to know the result can be proved in the $\sigma$-finite case, too. Thanks! $\endgroup$ – poppy3345 Aug 11 '15 at 18:54
  • $\begingroup$ I guess my one last question is whether we know automatically that $B = \{x : f(x) = 0\}$ is measurable? $\endgroup$ – poppy3345 Aug 11 '15 at 19:04
  • $\begingroup$ @gesa You can find the Radon-Nikodym for the case where both $\mu$ and $\sigma$-finite, for instance, in Measure Theory Halmos (Theorem 31.B). $\endgroup$ – Ramiro Aug 11 '15 at 19:12
  • $\begingroup$ @gesa $f$ is a measurable function. So, $B=\{x:f(x)=0\}$ is measurable. $\endgroup$ – Ramiro Aug 11 '15 at 19:14
  • $\begingroup$ Yep ok just forgot for a second that $f$ was measurable... thanks! $\endgroup$ – poppy3345 Aug 11 '15 at 19:16

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