Triangle Inequality for Integrals

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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

$\blacksquare$


Also see


Sources