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:


 * $\displaystyle \left\vert{\int_X f \, \mathrm d \mu}\right\vert \le \int_X \left\vert{f}\right\vert \, \mathrm d \mu$

Proof
Let $\displaystyle z = \int_X f \mathrm d \mu \in \C$.

By Complex Multiplication as Geometrical Transformation, there is a complex number $\alpha$ with $\left| \alpha \right| = 1$ such that $\alpha z = \left| z \right| \in \R$.

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 and Imaginary Part, we have that
 * $\displaystyle u \leq \left \vert \alpha f \right \vert = \left \vert f \right \vert$

Thus we get the inequality

Also see

 * Absolute Value of Complex Integral