Triangle Inequality for Integrals/Complex Function

Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $\struct {\C, \map \BB \C}$ be the complex numbers made into a measurable space with its Borel $\sigma$-algebra.

Then $\cmod f$ is $\mu$-integrable and:

$\ds \cmod {\int f \rd \mu} \le \int \cmod f \rd \mu$

Let $\struct {\R, \map \BB \R}$ be the real numbers made into a measurable space with its Borel $\sigma$-algebra.

We have:

$\ds \size {\map \Re f}^2 + \size {\map \Im f}^2$

$\le$

$\ds \size {\map \Re f}^2 + 2 \size {\map \Re f} \size {\map \Im f} + \size {\map \Im f}^2$

$\ds $

$=$

$\ds \paren {\size {\map \Re f} + \size {\map \Im f} }^2$

Taking square roots, we have:

$\cmod f \le \size {\map \Re f} + \size {\map \Im f}$

Since $f$ is $\mu$-integrable, $\map \Re f$ and $\map \Im f$ are $\mu$-integrable.

From Characterization of Integrable Functions, $\cmod {\map \Re f}$ and $\cmod {\map \Im f}$ are $\mu$-integrable.

Now, let:

$\ds z = \int f \rd \mu \in \C$

$\alpha z = \cmod z \in \R$

Let $u = \map \Re {\alpha f}$.

We have:

$\ds \alpha \int f \rd \mu = \int \alpha f \rd \mu$

Since:

$\ds \int \alpha f \rd \mu = \cmod {\int f \rd \mu} \in \R$

we have that:

$\ds \map \Im {\int \alpha f \rd \mu} = \int \map \Im {\alpha f} \rd \mu = 0$

Hence:

$\ds \int \alpha f \rd \mu = \map \Re {\int \alpha f \rd \mu} = \int \map \Re {\alpha f} \rd \mu = \int u \rd \mu$

$u \le \cmod {\alpha f} = \cmod f$

$\ds \int u \rd \mu \le \int \cmod f \rd \mu$

Putting everything together we have:

$\ds \cmod {\int f \rd \mu} \le \int \cmod f \rd \mu$

$\blacksquare$