R&C analysis - 1.9

Rudin's Real and Complex Analysis - Section 1.9 review:

Section $1.9$ deals with consequences of Theorem $1.9$ and $1.8$.
Taking $X$ as a measurable space, a function $f=u+iv$ is measurable when $u,v$ are measurable. This follows by setting
$\Phi(z)=\Phi(u,v)=z=u+iv$ where $\Phi$ is a continuous mapping of a plane into a topological space - $f$ is measurable.

Given $f=u+iv$ is measurable, showing $u$,$v$ are measurable is easy. Assume $Re(z)=u$ be projection of real part. Since $Re(z)$ is continuous, $Re\circ f$ is measurable via Theorem $1.7$. Similarly for $v$ and $|f|$.

Simple proof shows that $f+g$ and $fg$ are measurable for measurable $f,g$.

On a measurable set, definition of characterstic function is given. Then a theorem (point e) on complex measurable function is proved that has no applications for almost several sections!

The construction in this proof is instructive. Given that $f$ is a complex measurable function, need to show there exists a complex measurable function $\alpha$ such that $|\alpha|=1$ and $f=\alpha|f|$. Approach is to use Theorem $1.7$ that requires constructing a continuous function $\phi$ and compose $\phi$ with measurable functions.

Start off by setting $\phi(z)=z/|z|$ for each $z \in Y$ where $Y$ is complex plane with origin removed.

To use characterstic function - a set $E=\{x|f(x)=0\}$ is defined, then

$\alpha(x) = \phi(f(x)+\chi_E)$


Clearly, if $x \in E$, $\chi_E=1$ and $f(x)=0$. Then $\alpha(x)=1$.
If $x \notin E$, then $\chi_E=0$ and $\alpha(x)=\phi(f(x))=f(x)/|f(x)|$. Now Theorem $1.7$ can be applied to prove existence of $\alpha$.