Must two measures dominating another pair of measures be almost everywhere equivalent wrt. the dominated measures? I had a student who wrote the following, and I'm not sure whether this is correct or not in generality:
For probability measures $μ, ν, ρ, π$ defined on a general measure space, supposing

*

*$\rho \gg μ$,

*$ρ \gg ν$,

*$π \gg μ$, and

*$π \gg ν$;

there is a measurable set $A$ for which

*

*$ρ$ and $π$ restricted to $A$ are equivalent

*$μ(A) = 1 = ν(A)$
I tried and failed to provide a neat counterexample.
Can you help me find one, or otherwise convince me it is true?
 A: Since all measures here are probability measures, we can apply the Radon-Nikodym theorem to find measures $\lambda$ and $\kappa$ such that $\lambda \perp \rho$, $\kappa \ll \rho$, and $\pi = \lambda + \kappa$. By definition of mutual singularity, we can find $E$ such that $\lambda(E) = 0 = \rho(E^{c})$. Furthermore, $\mu, \nu \ll \rho$, so $\mu(E^{c}) = \nu(E^{c}) = 0$. Thus we can restrict our attention to $E$, where $\pi = \kappa \ll \rho$.
The previous paragraph allows us to reduce the problem to the original conditions plus $\pi \ll \rho$. We now want to find a set $A$ with $\mu(A) = 1 = \nu(A)$ such that $\rho \ll \pi$ on $A$. Let
$$
\mathcal{E} = \{E \subseteq X : \pi(E) = 0, \ \rho(E) > 0\},
$$
and assume this set is nonempty (otherwise we are done). We know that
$$
0 < \sup_{E \in \mathcal{E}} \rho(E) \leq 1,
$$
so let $E_{n} \in \mathcal{E}$ be such that
$$
\rho(E_{n}) \geq \sup_{E \in \mathcal{E}} \rho(E) - \frac{1}{n}.
$$
Define $F = \cup_{n = 1}^{\infty} E_{n}$. Then
$$
\rho(F) = \sup_{E \in \mathcal{E}} \rho(E_{n}) \quad \text{and} \quad \pi(F) = 0.
$$
Let $A = X \backslash F$. Since $\mu, \nu \ll \pi$, $\mu(A) = 1 = \nu(A)$. Moreover, for any $E \subseteq A$, if $\pi(E) = 0$, then
$$
\rho(E) + \rho(F) = \rho(E \cup F) \leq \rho(F)
$$
since $E \cup F \in \mathcal{E}$. Rearranging gives $\rho(E) = 0$, so $\rho \ll \pi$ on $A$.
A: Let $m = \rho + \pi$, so that all of $\rho, \pi, \mu, \nu$ are absolutely continuous to $m$.
Fix any representative of the Radon-Nikodym derivative $\frac{d\mu}{dm}$ and set $A_\mu = \{\frac{d\mu}{dm} > 0\}$.  Note that $\mu(A_\mu) = 1$ and that $m$ and $\mu$ are equivalent on $A_\mu$, i.e. for $E \subset A_\mu$ we have $\mu(E)=0$ iff $m(E)=0$.  Define $A_\nu$ similarly and let $A = A_\mu \cup A_\nu$, so that $\mu(A)=\nu(A)=1$.
Now suppose $E \subset A$ with $\rho(E) = 0$.  This forces $\mu(E)=0$ and in particular $\mu(E \cap A_\mu) = 0$, so $m(E \cap A_\mu) = 0$ and thus $\pi(E \cap A_\mu) = 0$.  Similarly, $\pi(E \cap A_\nu) = 0$, and we conclude $\pi(E)=0$.  Hence $\pi \ll \rho$ on $A$, and we similarly get $\rho \ll \pi$ on $A$ as well.
