Hahn–Kolmogorov theorem

In mathematics, the Hahn–Kolmogorov theorem characterizes when a finitely additive function with non-negative (possibly infinite) values can be extended to a bona fide measure. It is named after the Austrian mathematician Hans Hahn and the Russian/Soviet mathematician Andrey Kolmogorov.

Statement of the theorem

Let $\Sigma _{0}$ be an algebra of subsets of a set $X.$ Consider a function

$\mu _{0}\colon \Sigma _{0}\to \mathbb {R} \cup \{\infty \}$

which is finitely additive, meaning that

$\mu _{0}(\bigcup _{n=1}^{N}A_{n})=\sum _{n=1}^{N}\mu _{0}(A_{n})$

for any positive integer N and $A_{1},A_{2},\dots ,A_{N}$ disjoint sets in $\Sigma _{0}$.

Assume that this function satisfies the stronger sigma additivity assumption

$\mu _{0}(\bigcup _{n=1}^{\infty }A_{n})=\sum _{n=1}^{\infty }\mu _{0}(A_{n})$

for any disjoint family $\{A_{n}:n\in \mathbb {N} \}$ of elements of $\Sigma _{0}$ such that $\cup _{n=1}^{\infty }A_{n}\in \Sigma _{0}$. (Functions $\mu _{0}$ obeying these two properties are known as pre-measures.) Then, $\mu _{0}$ extends to a measure defined on the sigma-algebra $\Sigma$ generated by $\Sigma _{0}$; i.e., there exists a measure

$\mu \colon \Sigma \to \mathbb {R} \cup \{\infty \}$

such that its restriction to $\Sigma _{0}$ coincides with $\mu _{0}.$

If $\mu _{0}$ is $\sigma$-finite, then the extension is unique.

Non-uniqueness of the extension

If $\mu _{0}$ is not $\sigma$-finite then the extension need not be unique, even if the extension itself is $\sigma$-finite.

Here is an example:

We call rational closed-open interval, any subset of $\mathbb {Q}$ of the form $[a,b)$, where $a,b\in \mathbb {Q}$.

Let $X$ be $\mathbb {Q} \cap [0,1)$ and let $\Sigma _{0}$ be the algebra of all finite union of rational closed-open intervals contained in $\mathbb {Q} \cap [0,1)$. It is easy to prove that $\Sigma _{0}$ is, in fact, an algebra. It is also easy to see that every non-empty set in $\Sigma _{0}$ is infinite.

Let $\mu _{0}$ be the counting set function ($\#$) defined in $\Sigma _{0}$. It is clear that $\mu _{0}$ is finitely additive and $\sigma$-additive in $\Sigma _{0}$. Since every non-empty set in $\Sigma _{0}$ is infinite, we have, for every non-empty set $A\in \Sigma _{0}$, $\mu _{0}(A)=+\infty$

Now, let $\Sigma$ be the $\sigma$-algebra generated by $\Sigma _{0}$. It is easy to see that $\Sigma$ is the set of all subset of $X$, and both $\#$ and $2.\#$ are $\sigma$-finite measures defined on $\Sigma$ and both are extensions of $\mu _{0}$.

Comments

This theorem is remarkable for it allows one to construct a measure by first defining it on a small algebra of sets, where its sigma additivity could be easy to verify, and then this theorem guarantees its extension to a sigma-algebra. The proof of this theorem is not trivial, since it requires extending $\mu _{0}$ from an algebra of sets to a potentially much bigger sigma-algebra, guaranteeing that the extension is unique (if $\mu _{0}$ is $\sigma$-finite), and moreover that it does not fail to satisfy the sigma-additivity of the original function.

See also

• Carathéodory's extension theorem
• pre-measure

---

Hahn–Kolmogorov theorem at PlanetMath.