Let $\mu,\nu$ be two finite positive measures. Show that, there is a set $E$ such that $\mu_E\ll\nu_E$ and $\mu_{E^c}\perp\nu_{E^c}$

Question

Let $\mu,\nu$ be two finite positive measures on $(X,\mathfrak{M})$. For $E\in\mathfrak{M}$, we define $\mu_E(A):=\mu(A\cap E)$ and similarly $\nu_E$. Show that, there is $E\in\mathfrak{M}$ such that $\mu_E\ll\nu_E$ and $\mu_{X\setminus E}\perp\nu_{X\setminus E}$

We define the signed measure $\lambda=\mu-\nu$. By John Hahn Decompostition, $\lambda=\lambda^+-\lambda^-$ and there are $E,F\in\mathfrak{M}$ which are disjoint, $E\cup F=X$ such that $\lambda^+=\lambda_E$ and $\lambda^-=\lambda_F$.

So, $\lambda^+=\mu_E-\nu_E$ and $\lambda^-=\mu_F-\nu_F$. We have $\lambda^+\perp\lambda^-\implies \mu_E-\nu_E\perp \mu_F-\nu_F\implies \mu_E\perp\nu_F$ and $\mu_F\perp\nu_E$. But this is obvious, because $\lambda_E$ and $\nu_F$ are concentrated on $E$ and $F$ respectively and they are disjoint. I have to find $E$ such that $\mu_E\perp\nu_E$.

And I am not having any way out for the absolute continuity case? Can anyone help me to solve the problem?

Thanks for your help in advance.

Answer 1

Hint: Let $\lambda =\mu +\nu$. Be Radon Nikodym Theorem we can write $\mu(A)=\int_A f d\lambda$ and $\nu(A)=\int_A g d\lambda$. Let $E=\{x: f(x)g(x)>0\}=\{x: f(x)>0\}\cap \{x: g(x)>0\}$ and $F=\{x: f(x)g(x)=0\}$. It is fairly easy to see that this meets the requirements.