Triangle Inequality for Integrals/Proof 1

Proof
Let $\ds z = \int_X f \rd \mu \in \C$.

By Complex Multiplication as Geometrical Transformation, there is a complex number $\alpha$ with $\cmod \alpha = 1$ such that:
 * $\alpha z = \cmod z \in \R$

Let $u = \map \Re {\alpha f}$, where $\Re$ denotes the real part of a complex number.

By Modulus Larger than Real Part, we have that:
 * $u \le \cmod {\alpha f} = \cmod f$

Thus we get the inequality: