---
tag: CM6XF84MNY
kind: definition
name: Loeb measure
msc: ["28E05", "26E35"]
defines: DW5SBZXK6R
uses: [PR11BZQQPG, P15VQS5DJ9, KFWQ24D9M6]
sources:
  - url: https://en.wikipedia.org/wiki/Loeb_space
---

## Definition

Work in a nonstandard extension of the real numbers that is countably saturated.
Let $\Omega$ be an internal set and let $\mathcal{A}$ be an internal algebra of subsets of $\Omega$.
Let $\mu \colon \mathcal{A} \to {}^*[0, \infty)$ be an internal [[finitely additive measure|KFWQ24D9M6]] such that $\mu(\Omega)$ is a finite hyperreal number.
For $A \in \mathcal{A}$ let $\operatorname{st}\mu(A)$ be the standard part of $\mu(A)$, that is, the unique real number infinitely close to $\mu(A)$.
The *Loeb measure* $L_\mu$ of $\mu$ is the unique [[measure|PR11BZQQPG]] on the σ-algebra $\sigma(\mathcal{A})$ generated by $\mathcal{A}$ such that $L_\mu(A) = \operatorname{st}\mu(A)$ for all $A \in \mathcal{A}$.
The [[measure space|P15VQS5DJ9]] $(\Omega, \sigma(\mathcal{A}), L_\mu)$ is the *Loeb space* of $\mu$.

## Remarks

The map $\operatorname{st}\mu$ is finitely additive on $\mathcal{A}$.
By countable saturation, if a countable union of sets of $\mathcal{A}$ lies in $\mathcal{A}$, then it is already a finite union, so $\operatorname{st}\mu$ is countably additive on $\mathcal{A}$.
Hence [[Carathéodory's extension theorem|Q8EXG7EKYW]] gives the extension, and it is unique because $\operatorname{st}\mu(\Omega)$ is finite.
Many authors call the [[completion|KJNSKV75VQ]] of this measure the Loeb measure.

Example: let $N$ be an infinite hypernatural number, let $\Omega = \set{1, \ldots, N}$, and let $\mathcal{A}$ be the internal power set of $\Omega$.
Let $\mu(A) = \abs{A}/N$, where $\abs{A}$ is the internal number of elements of $A$.
Then $\mu(\Omega) = 1$ is finite, and $L_\mu$ is a probability measure.
The standard part map $k \mapsto \operatorname{st}(k/N)$ pushes $L_\mu$ forward to Lebesgue measure on $[0, 1]$.

Non-example: on the same $\Omega$, the internal counting measure $\mu(A) = \abs{A}$ has $\mu(\Omega) = N$, which is infinite.
It fails the finiteness clause, and the construction above does not apply directly.
For such $\mu$ one can still define a Loeb measure, through an outer measure, but it need not be the unique extension.