In Folland's "Real Analysis", two complex measures $\nu=\nu_r + i\nu_i$ and $\mu=\mu_r + i\mu_i$ are said to be mutually singular (in symbols: $\nu\perp \mu$) if "$\nu_a \perp \mu_b$ for $a,b = r,i$". While I do understand, what it means for two signed measures to be mutually singular (Folland explains this clearly), the content and meaning of this definition is not particularly clear to me.
I guess the definition should be read this way: $\nu \perp \mu$ (by definition) if $\nu_r \perp \mu_r$ and $\nu_i \perp \mu_i$. [Edit: This guess is wrong, see the comments below]. The problem is, that what we seem to need in the discussion just below the definition is something (seemingly) stronger, namely something like this: $\nu \perp \mu$ iff the measurable space $X$, on which $\nu$ an $\mu$ are defined, is the disjoint union of two measurable sets, $E$ and $F$, such that $\nu(E)=0$ and $\mu(F)=0$.
To be more specific, we want to use Theorem 3.12 (The Lebesgue-Radon-Nikodym Theorem for complex measures) to argue that if $\nu$ is a complex measure on $X$, then $d\nu=\int f d \mu$ for some finite measure $\mu$ and some $f \in L^1(\mu)$. I know how to prove this using Lebesgue-Radon-Nikodym for signed measures (Theorem 3.8 in Folland), but the book suggests that the reader should use Theorem 3.12.
The obvious way to argue using Theorem 3.12 seems to be something like this: Given a complex measure $\nu$, let $\mu =|\nu_r|+|\nu_i|$. Then $\mu$ is finite, and $\nu \ll \mu$. Therefore, according to Theorem 3.12, there exists a complex measure $\lambda$ and some $f \in L^1(\mu)$ such that $\lambda \perp \mu$ and $d\nu=d\lambda + f d\mu$. Clearly, if we can show that $\lambda=0$, we are done. This follows quite easily, if "$\lambda \perp \mu$" is taken in the strong sense, I explain above. But does it also follow if we use Folland's definition of "$\lambda \perp \mu$"? Maybe the two definitions are equivalent in some way that I just don't get.
How does this look?
Definition: Two complex measures $\nu=\nu_r+i\nu_i$ and $μ=μ_r+iμ_i$ are said to be mutually singular (in which case we write ν⊥μ) if $\nu_r \perp \mu_r$, $\nu_r \perp \mu_i$, $\nu_i \perp \mu_r$ and $\nu_i \perp \mu_i$.
Proposition: Two complex measures $\nu$ and $\mu$ on a measurable space $(X, M)$ are mutually singular if and only if there exists disjoint measurable sets $J$ and $K$ such that $X=J \cup K$, $\nu(J)=0$ and $\mu(K)=0$.
Proof: The "if" part is an easy check, so we only prove "only if". So let us assume that $\nu \perp \mu$ and show that this implies the existence of $J$ and $K$ as in the statement. We begin by choosing measureable sets $A, B, C, D, E, F, G$ and $H$ such that:
$X=A \cup B$ , $A \cap B=\emptyset$, $A$ is null for $\nu_r$, $B$ is null for $\mu_r$
$X=C \cup D$ , $C \cap D=\emptyset$, $C$ is null for $\nu_r$, $D$ is null for $\mu_i$.
$X=E \cup F$ , $E \cap F=\emptyset$, $E$ is null for $\nu_i$, $F$ is null for $\mu_r$.
$X=G \cup H$ , $G \cap H=\emptyset$, $G$ is null for $\nu_i$, $H$ is null for $\mu_i$.
(Remark: This is possible by the definition of two signed measures being mutually singular.)
Now, if we let $J:=(A \cup C) \cap (E \cup G)$ and $K:=\complement J =(B\cap D)\cup (F \cap H)$ then $X=J \cup K$, $J \cap K = \emptyset$, $\nu(J)=0$ and $\mu(K)=0$.