Triangle Inequality for Integrals

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

Let $f: X \to \overline \R$ be a $\mu$-integrable function.

Then:


 * $\ds \size {\int_X f \rd \mu} \le \int_X \size f \rd \mu$

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:

Also see

 * Modulus of Complex Integral