# Triangle Inequality for Integrals

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## Contents

## 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

- 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 $1.33$