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:

\[ \int f d\mu = \int \re(f)d\mu + \int \im(f) d\mu . \]

A linear combination
of two integrable complex-valued
functions is an integrable
complex-valued function.

The integral is a linear
operator on the vector space
of integrable complex-valued
functions.

The absolute value of the integral of a
complex-valued function is smaller than the
integral of the absolute value of the function.

Let $(X, \mathcal{A} , \mu )$ be a measure
space.
Let $f: X \to \C $ integrable.
There exists $\alpha \in C$ with
$\Cmod{\alpha } = 1$ such that

\[ \Cmod{\int f d\mu } = \alpha \int f d\mu . \]

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 $\Cmod{\alpha } = 1$.- Future editions will modify this motivation. ↩︎