Triangle Inequality for Integrals/Proof 1
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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:
- $\ds \size {\int_X f \rd \mu} \le \int_X \size f \rd \mu$
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:
\(\ds \cmod {\int_X f \rd \mu}\) | \(=\) | \(\ds \alpha \int_X f \rd \mu\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_X \alpha f \rd \mu\) | Integral of Integrable Function is Homogeneous | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_X u \rd \mu\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \int_X \cmod f \rd \mu\) | Integral of Integrable Function is Monotone |
$\blacksquare$