2
$\begingroup$

I was reading Rudin and I stumbled upon a proof that I do not seem to understand. It is on page 325 of Baby Rudin $3^{rd}$ edition.

In case you do not have a copy I shall write some background information:

Suppose $\mu$ is a measure on $X$, and $f$ is a complex measurable function on $X$. Then, $\int |f|\;d\mu < +\infty$ and we define $\int f \;d\mu = \int u \; d\mu + i\int v \; d\mu$.

Now onto, my question. He wants to prove the following $ \left| \int f \; d\mu \right| \leq \int |f| \; d\mu. $

He begins as follows:

If $f \in \mathscr{L}(\mu)$, there is a complex number $c$, $|c| =1$, such that $ c\int f \; d\mu \geq 0 $

Put $g = cf = u +iv$ where $u$ and $v$ are real.Then

$ \left| \int f \; d\mu \right| = c\int f \;d\mu = \int g \;d\mu = \int u \;d\mu \leq \int |f| \; d\mu.$

The thing that bothers me is the following equality $\int g \;d\mu = \int u\;d\mu .$

How are these two functions equal? If we had assumed that $g = u +iv$, so should not g be $\int g \;d\mu = \int u \; d\mu + i\int v \; d\mu?$.

  • 0
    @AmiteshDatta Cool, thank you very much!2012-04-09

1 Answers 1

3

\left | \int f d \mu \right | is real so $ \left| \int f \; d\mu \right| = \int u \; d\mu + i\int v \; d\mu$

has to be real too, which means that $\int v d \mu $ has to vanish.