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

$\mu_0\colon \Sigma_0 \to[0,\infty]$

which is finitely additive, meaning that

$\mu_0\left(\bigcup_{n=1}^N A_n\right)=\sum_{n=1}^N \mu_0(A_n)$

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

$\mu_0\left(\bigcup_{n=1}^\infty A_n\right) = \sum_{n=1}^\infty \mu_0(A_n)$

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

$\mu \colon \Sigma \to[0,\infty]$

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 Borel $\sigma$-algebra of subsets of $X$, and both $\#$ and $2\#$ are measures defined on $\Sigma$ and both are extensions of $\mu_0$.

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.