We extend integrability to complex functions.1
A measurable complex-valued function on a measurable space is one whose real and imaginary parts are both measurable. An integrable complex-valued function on a measurable space is one whose real and imaginary parts are both integrable.
The integral of a integrable complex-valued function on a measurable space is the complex number whose real part is the integral of the real part of the function and whose imaginary part is the integral imaginary part of the function.
Let $(X, \mathcal{A} , \mu )$
be a measure space.
Let $C$ denote the set
of complex numbers.
Let $f: X \to C$
be a function.
$f$ is measurable
if $\re(f)$ and $\im(f)$
are measurable.
$f$ is integrable if
$\re(f)$ and $\im(f)$
is integrable.
If $f$ is integrable,
we denote its integral
by $\int f d\mu $.
We have defined it by:
\[
\int f d\mu = \int \re(f)d\mu + \int \im(f) d\mu .
\]
\[ \Cmod{\int f d\mu } = \alpha \int f d\mu . \]
Since the integral is homogenous,\[ \begin{aligned} \Cmod{\int f d\mu } &= \int \alpha f d\mu &= \int \re(\alpha f) d\mu + i \int \im(\alpha f)d\mu . \end{aligned} \]
Since $\C mod{\int f d\mu }$ is real, $\int \im(\alpha f) d\mu = 0$, so\[ \begin{aligned} \Cmod{\int f d\mu } &= \int \re(\alpha f) d\mu &\leq \int \abs{\alpha f} d\mu &= \int \Cmod{f} d\mu , \end{aligned} \]
since $\re(z) \leq \Cmod{z}$ for all complex numbers $z$ and $\Cmod{\alpha } = 1$.