# 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:

$\displaystyle \size {\int_X f \rd \mu} \le \int_X \size f \rd \mu$

## Proof

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

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:

 $\displaystyle \cmod {\int_X f \rd \mu}$ $=$ $\displaystyle \alpha \int_X f \rd \mu$ $\displaystyle$ $=$ $\displaystyle \int_X \alpha f \rd \mu$ Integral of Integrable Function is Homogeneous $\displaystyle$ $=$ $\displaystyle \int_X u \rd \mu$ $\displaystyle$ $\le$ $\displaystyle \int_X \cmod f \rd \mu$ Integral of Integrable Function is Monotone

$\blacksquare$