Theorem regarding the total variation measure My question relates to the proof of Theorem 6.13 (Chapter 6, Complex Measures, page 125) from Walter Rudin's Real & Complex Analysis (Third Edition).
Given:
$ds = g dp$ and
$ds = h d|s|$
Where:


*

*$p$ is a positive measure on a $\sigma$-algebra $\mathfrak M$

*$g$ and $h$ belong to $L^1(p)$

*$|s|$ is the total variation measure corresponding to the measure $s$

*$|h| = 1$
How does Rudin conclude that:
$d|s| = \bar{h} g ~dp$ 
Any thoughts will be helpful. 
Thanks.
 A: *

*Set the two expressions equal to one another:


$g \mathrm{d}p = h\mathrm{d} |s| $


*Then multiply both sides by $h'$:


$h' g \mathrm{d}p = |h| \mathrm{d} |s| $


*simplify using the fact that |h|=1:


$h' g \mathrm{d}p = \mathrm{d} |s| $
Edit: justification of the above steps
$ g \mathrm{d}p = h\mathrm{d} |s| $  as measures, so $ \int_E f g \mathrm{d}p = \int_E f h\mathrm{d} |s| $ for every measurable function $f$ and every measurable set $E$. In particular, for $f = h'$, $ \int_E h' g \mathrm{d}p = \int_E h' h\mathrm{d} |s| $. Therefore,  $ \int_E h' g \mathrm{d}p = \int_E \mathrm{d} |s| $ for every measurable $E$. Hence $h' g \mathrm{d}p = \mathrm{d} |s| $.
A: I was able to solve the problem by decomposing the complex measures "gdp" and "hd|s|" into 4 positive measures (each) and then applying theorem 1.29 to each of the positive measures and h' (by splitting h' into 4 non negative measurable functions using h' = u1 - u2 + i(v1 - v2) where u1, u2, v1 and v2 are real and non negative.
Thanks for the comments.
