I am stuck in the proof of Theorem 6.13 in rudin's real and complex analysis,First i quote the proof briefly "there is a function $h$, of absolute value 1, such that $d\lambda=h\ d|\lambda|$. And by hypothesis $d\lambda=g\ d\mu$. Hence,
$h\ d|\lambda|=g\ d\mu
$
This gives $d|\lambda|=\bar{h}g\ d\mu$ ,where lambda is a complex measure and mu is a positive sigma finite measure"
\I am stuck in the last line that how to show $d|\lambda|=\bar{h}g\ d\mu$ from $h\ d|\lambda|=g\ d\mu$. any help will be appreciated.
according to the meaning of notation,$$d \lambda = h d |\lambda | \implies \lambda (E)= \int _E h d | \lambda |$$ for all $ E \in M$ and
$d\lambda = g dx \implies \lambda (E) = \int _E g dx$ for all $E \in M$.
Now
$\int E h d | \lambda | = \int _E gdx$ for all $ E \in M$,now
from here how can I show $| \lambda | (E) = \int _E \overline h g dx $ for all $E\in M$?
Multiply both sides by $\bar{h}$ and note that $h\bar{h} = |h|^2 = 1$ by assumption.
More details in response to comment:
We are given that $h \, d|\lambda| = g\, d\mu$. The definition of this is $$\int \chi_E h \, d|\lambda| = \int \chi_E g \, d\mu, \qquad \forall E \in M,$$ where $\chi_E$ is the indicator function for $E$.
If a function $r$ is measurable, then by approximating it with simple functions we also have $$\int rh \, d|\lambda| = \int rg \, d\mu.$$
In particular if we let $r = \chi_E \bar{h}$ we obtain $$\int \chi_E \bar{h} h \, d|\lambda| = \int \chi_E \bar{h} g \, d\mu.$$
The left-hand side is $|\lambda|(E)$ and the right-hand side is $\int_E \bar{h}g \, d\mu$. Since this holds for all $E \in M$, we have $d|\lambda| = \bar{h}g\, d\mu$.