Is $\nu_1\perp\nu_2$ equivalent to $|\nu_1|\perp|\nu_2|$ for complex measure? Suppose $\nu_1$ and $\nu_2$ are two complex measures, do we have $\nu_1\perp\nu_2\Leftrightarrow|\nu_1|\perp|\nu_2|$? if yes, please give a proof. If no, please give a counterexample. Thanks!
PS: two complex measure $\nu_1\perp\nu_2$ if so are their real and imaginary parts, respectively.
 A: I solved it by myself.
It is easy to prove $|\nu_1|\perp|\nu_2|\Rightarrow\nu_1\perp\nu_2$ so I omit it. Now we prove the converse.
Suppose measurable sets $E_1$ and $F_1$ makes $\nu_{1r}\perp\nu_{2r}$, that is, $E_1\cap F_1=\emptyset, E_1\cup F_1=X$, $E_1$ is null for $\nu_{1r}$ and $F_1$ is null for $\nu_{2r}$. For $\nu_{1r}\perp\nu_{2i}$, suppose $E_2$ and $F_2$ makes this mutual singularity. Since $\nu_{1r}$ remains the same in these two situations, $E_1\cup E_2$ is null for $\nu_{1r}$ because by subadditivity $|\nu_{1r}|(E_1\cup E_2)\leq|\nu_{1r}|(E_1)+|\nu_{1r}|(E_2)=0$. While on the other hand its complement $F_1\cap F_2$, as a subset of $F_1$, is null for $\nu_{2r}$, Also it is null for $\nu_{2i}$ for the same reason, so $F_1\cap F_2$ is null for $\nu_2$. Supposing $E_3$ and $F_3$ makes $\nu_{1i}\perp\nu_{2r}$ and $E_4$ and $F_4$ makes $\nu_{1i}\perp\nu_{2i}$, the above argument produces $E_3\cup E_4$ is null for $\nu_{1i}$ and $F_3\cap F_4$ is null for $\nu_2$. Now it is $\nu_2$ that remains the same. So we construct $E=(E_1\cup E_2)\cap(E_3\cup E_4), F=E^c=(F_1\cap F_2)\cup(F_3\cap F_4)$. Using the same reasoning we can verify that E is null for both $\nu_{1r}$ and $\nu_{1i}$, so it is null for $\nu_1$. And F is null for $\nu_2$ by subadditivity.
