Real and Imaginary parts of Lebesgue integral in L1 (folland's real analysis)

Question

I am working through Folland's Real Analysis and have a question about a proposition on functions in L1. Prop (2.22) states:

given $f \in L^1$, then $|\int f| \leq \int|f|$

In the proof, for when f is complex valued, they define

$ \alpha = sgn(\int f)$ and state that $\int \alpha f$ is real. I am unsure about how this statement is true. My understanding is that if $f$ is complex, then $\int f$ is the supremum of simple functions with complex coefficients,

i.e. $a_j \in \mathbb{C}$, and

$\int f = sup\{\int \phi\::\: 0 \leq \phi \leq f,\: \phi\:simple \}$ where

$\int \phi = \sum a_j \mu(E_j)$

Wouldn't this make the integral complex?

Thanks for any help!

Answer 1

Folland is using the complex sign, defined by $sgn(x) = \frac{x}{|x|}$

$\alpha = \frac{\overline{\int f}}{|\int f|}$, so $\alpha \int f$ is real.

Answer 2

Recall that for Lebesgue integrable functions, $\int c\cdot f=c\cdot \int f$. Now, $\alpha$ is chosen so that $\alpha \cdot \int f$ is real, and since $\int \alpha f=\alpha \int f$ it follows that $\int \alpha f$ is real.