Triangle Inequality for Integrals

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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

$\blacksquare$


Also see


Sources