Htykuut

This is an exercise from Folland. It is exercise 3.19

The question: Prove for complex measures $mu$ and $nu$ that $nuperp mu$ iff $vert nu vertperp vert mu vert$.

The $(Leftarrow)$ direction is a very simple consequence of the fact that $vert nu(E)vertleq vert nu vert (E)$ 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 $dnu=fdmu$ for $mu$ a positive measure, then $dvert nu vert=vert f vert dmu$

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

$nu_iperp mu_i$

$nu_iperp mu_r$

$nu_r perp mu_i$

$nu_r perp mu_r$

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.

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 $(f cdot rho)(mathrm{X}) = intlimits_{mathrm{X}} f drho$ 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.

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 $|mu| perp |nu|$.