# Rudin Question (Integration of Complex Functions) [pg.325]

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 $$\int g \;d\mu = \int u \; d\mu + i\int v \; d\mu?$$.

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$\left | \int f d \mu \right |$ is real so $\int v$ has to be zero otherwise you get something complex on the RHS. – Rudy the Reindeer Apr 9 '12 at 7:23
What's $\mathscr L (\mu)$? – Rudy the Reindeer Apr 9 '12 at 7:24
@MattN. If I remember correctly from my reading of Rudin's textbook, then I believe that $\cal L(\mu)$ is the $L^1$-space of the positive measure $\mu$. I think Rudin's notation is not standard in this regard. – Amitesh Datta Apr 9 '12 at 13:21
@AmiteshDatta Cool, thank you very much! – Rudy the Reindeer Apr 9 '12 at 13:30

$$\left | \int f d \mu \right |$$ 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.