Mutually singular complex measures implies mutual singularity of total variation

Question

This is an exercise from Folland. It is exercise 3.19

The question: Prove for complex measures $\mu$ and $\nu$ that $\nu\perp \mu$ iff $\vert \nu \vert\perp \vert \mu \vert$.

The $(\Leftarrow)$ direction is a very simple consequence of the fact that $\vert \nu(E)\vert\leq \vert \nu \vert (E)$ for any complex measure $\nu$. The other direction is the direction I am struggling with.

For a complex measure $\nu$, I am using the definition of $\vert \nu \vert$ which says $\vert \nu \vert$ is the measure which is determined by if $d\nu=fd\mu$ for $\mu$ a positive measure, then $d\vert \nu \vert=\vert f \vert d\mu$

There doesn't seem to be a lot we can use. First we see, by definition, that $\nu \perp \mu$ gives rise to the following

$\nu_i\perp \mu_i$

$\nu_i\perp \mu_r$

$\nu_r \perp \mu_i$

$\nu_r \perp \mu_r$

Though this doesn't give me a lot to work with, as I don't have a way of expressing $\vert \nu \vert$ in terms of $\nu_i$ and $\nu_r$...I would prefer a hint to a complete answer.

Answer 1

Full solution. Write $\nu = f \cdot \rho$ and $\mu = g \cdot \sigma,$ where $\rho$ and $\sigma$ are positive measures, here I use the common notation $(f \cdot \rho)(\mathrm{X}) = \int\limits_{\mathrm{X}} f\ d\rho$ and similar for other cases. Consider the positive measure $\tau = \rho + \sigma$ and the Lebesgue-Nikodym theorem asserts that $\rho \ll \tau$ and $\sigma \ll \tau$ so that $\nu = f_0 \cdot \tau$ and $\mu = g_0 \cdot \tau.$ Then, $|\nu| = |f_0| \cdot \tau$ and $|\mu| = |g_0| \cdot \tau.$ the hypothesis $\nu \perp \mu$ signifies then $f_0 g_0 = 0$ for almost every point with respect to $\tau,$ hence $|f_0| |g_0| = 0$ for almost every point with respect to $\tau,$ and this is the same as saying $|\nu| \perp |\mu|.$ Q.E.D.

Answer 2

I don't have Folland's book and I will use the definitions in Rudin's book. $\mu \perp \nu$ means there exist disjoint measurable sets $A,B$ such that $\mu$ is concentrated on $A$ and $\nu$ is concentrated on $B$. $\mu$ is concentrated on $A$ and means $\mu (E)=0$ for every measurable set $E \subset A^{c}$. Hence $|\mu| (E) =\sup \{\sum |\mu (E_n)|: \{E_n\} \text {is a partition of } E\}=0$ for every measurable set $E \subset A^{c}$. (Because $\mu (E_n)=0$ for every $n$). Hence $|\mu |(A^{c})=0$. Similarly, $|\nu |(B^{c})=0$. Hence $|\mu| \perp |\nu|$.