Hahn–Kolmogorov theorem

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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[edit]

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[edit]

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.

Comments[edit]

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[edit]

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