# Triangle Inequality for Integrals

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## Contents

## 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

- 2005: René L. Schilling:
*Measures, Integrals and Martingales*... (previous) ... (next): $10.4 \ \text{(v)}$ - 1966: Walter Rudin:
*Real and Complex Analysis*: $1$: Integration of complex functions: Theorem $1.33$