$\|h\mu\|= \int |h|d|\mu|$

Question

I was reading Conway and just popped in. I am trying to prove the note that part, but I cannot. Any help will be appreciated.

Answer 1

There exists a measurable function $\phi$ such that $|\phi|=1$ a.e. [$|\mu|$] and $d\mu =\phi d|\mu|$. Let $\nu(E)=\int_E h\phi d|\mu|$ so $\nu(E)=\int_E h d\mu$. It is clear from definition that $|\nu| \leq \int |h\phi | d|\mu|=\int |h | d|\mu|$. Now show that $\int_E|h|d|\mu|=\int_E \overline {\phi} g d\nu$ where $g=\frac h {|h|}$ when $h \neq 0$ and $1$ when $h=0$. [ Use simple function approximation]. This gives $\int |h|d|\mu| \leq |\nu|$.