Hahn–Kolmogorov theorem

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

${\displaystyle \mu _{0}\colon \Sigma _{0}\to [0,\infty ]}$

which is finitely additive, meaning that

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

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

Assume that this function satisfies the stronger sigma additivity assumption

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

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

${\displaystyle \mu \colon \Sigma \to [0,\infty ]}$

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

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

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

Here is an example:

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

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

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

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

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 ${\displaystyle \mu _{0}}$ from an algebra of sets to a potentially much bigger sigma-algebra, guaranteeing that the extension is unique (if ${\displaystyle \mu _{0}}$ is ${\displaystyle \sigma }$-finite), and moreover that it does not fail to satisfy the sigma-additivity of the original function.

• Carathéodory's extension theorem
• pre-measure

This article incorporates material from Hahn–Kolmogorov theorem on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.