# Triangle Inequality for Integrals

## Theorem

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

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

Then:

$\displaystyle \left\vert{\int_X f \rd \mu}\right\vert \le \int_X \left\vert{f}\right\vert \rd \mu$

## Proof

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

By Complex Multiplication as Geometrical Transformation, there is a complex number $\alpha$ with $\left\vert{\alpha}\right\vert = 1$ such that:

$\alpha z = \left\vert{z}\right\vert \in \R$

Let $u = \mathfrak{Re} \left({\alpha f}\right)$, where $\mathfrak{Re}$ denotes the real part of a complex number.

By Modulus Larger than Real Part, we have that:

$\displaystyle u \le \left\vert{\alpha f}\right\vert = \left\vert{f}\right\vert$

Thus we get the inequality:

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

$\blacksquare$