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 Russia n/Soviet mathematician Andrey Kolmogorov.

Statement of the theorem

Let ${\mathrm{\Sigma }}_0$ be an algebra of subsets of a set $X.$ Consider a function

${\mu }_0:{\mathrm{\Sigma }}_0\to [0,\mathrm{\infty }]$

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 ${\mathrm{\Sigma }}_0$.

Assume that this function satisfies the stronger sigma additivity assumption

${\mu }_0\left(\bigcup _{n=1}^{\mathrm{\infty }}{A}_n\right)=\sum _{n=1}^{\mathrm{\infty }}{\mu }_0(A_n)$

for any disjoint family $\{A_n:n\in \mathbb{\mathbb{N}}\}$ of elements of ${\mathrm{\Sigma }}_0$ such that $\underset{n=1}{\overset{\mathrm{\infty }}{\cup }}{A}_n\in {\mathrm{\Sigma }}_0$. (Functions ${\mu }_0$ obeying these two properties are known as pre-measures.) Then, ${\mu }_0$ extends to a measure defined on the sigma-algebra $\mathrm{\Sigma }$ generated by ${\mathrm{\Sigma }}_0$; i.e., there exists a measure

$\mu :\mathrm{\Sigma }\to [0,\mathrm{\infty }]$

such that its restriction to ${\mathrm{\Sigma }}_0$ 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 $\mathbb{\mathbb{Q}}$ of the form $[a,b)$, where $a,b\in \mathbb{\mathbb{Q}}$.

Let $X$ be $\mathbb{\mathbb{Q}}\cap [0,1)$ and let ${\mathrm{\Sigma }}_0$ be the algebra of all finite union of rational closed-open intervals contained in $\mathbb{\mathbb{Q}}\cap [0,1)$. It is easy to prove that ${\mathrm{\Sigma }}_0$ is, in fact, an algebra. It is also easy to see that the cardinal of every non-empty set in ${\mathrm{\Sigma }}_0$ is ${\mathrm{\aleph }}_0$.

Let ${\mu }_0$ be the counting set function ($\mathrm{\#}$) defined in ${\mathrm{\Sigma }}_0$. It is clear that ${\mu }_0$ is finitely additive and $\sigma$-additive in ${\mathrm{\Sigma }}_0$. Since every non-empty set in ${\mathrm{\Sigma }}_0$ is infinite, we have, for every non-empty set $A\in {\mathrm{\Sigma }}_0$, ${\mu }_0(A)=+\mathrm{\infty }$

Now, let $\mathrm{\Sigma }$ be the $\sigma$-algebra generated by ${\mathrm{\Sigma }}_0$. It is easy to see that $\mathrm{\Sigma }$ is the Borel $\sigma$-algebra of subsets of $X$, and both $\mathrm{\#}$ and $2\mathrm{\#}$ are measures defined on $\mathrm{\Sigma }$ 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