Carathéodory's extension theorem#Introductory statement

Several very similar statements of the theorem can be given. A slightly more involved one, based on semi-rings of sets, is given further down below. A shorter, simpler statement is as follows. In this form, it is often called the Hahn–Kolmogorov theorem.

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

$$\backslash mu\_0\; :\; \backslash Sigma\_0\; \backslash to\; [0,\; \backslash infty]$$

which is sigma additive, meaning that

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

for any disjoint family $\backslash \{A\_n\; :\; n\; \backslash in\; \backslash N\backslash \}$ of elements of $\backslash Sigma\_0$ such that $\backslash cup\_\{n=1\}^\backslash infty\; A\_n\; \backslash in\; \backslash Sigma\_0.$ (Functions $\backslash mu\_0$ obeying these two properties are known as pre-measures.) Then,

$\backslash mu\_0$ extends to a measure defined on the $\backslash sigma$-algebra $\backslash Sigma$ generated by $\backslash Sigma\_0$; that is, there exists a measure

$$\backslash mu\; :\; \backslash Sigma\; \backslash to\; [0,\; \backslash infty]$$

such that its restriction to $\backslash Sigma\_0$ coincides with $\backslash mu\_0.$

If $\backslash mu\_0$ is $\backslash sigma$-finite, then the extension is unique.

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 $\backslash mu\_0$ from an algebra of sets to a potentially much bigger sigma-algebra, guaranteeing that the extension is unique (if $\backslash mu\_0$ is $\backslash sigma$-finite), and moreover that it does not fail to satisfy the sigma-additivity of the original function.

For a given set $\backslash Omega,$ we call a family $\backslash mathcal\{S\}$ of subsets of $\backslash Omega$ a Semi-ring of sets if it has the following properties:

• $\backslash varnothing\; \backslash in\; \backslash mathcal\{S\}$
• For all $A,\; B\; \backslash in\; \backslash mathcal\{S\},$ we have $A\; \backslash cap\; B\; \backslash in\; \backslash mathcal\{S\}$ (closed under pairwise intersections)
• For all $A,\; B\; \backslash in\; \backslash mathcal\{S\},$ there exists a finite number of disjoint sets $K\_i\; \backslash in\; \backslash mathcal\{S\},\; i\; =\; 1,\; 2,\; \backslash ldots,\; n,$ such that $A\; \backslash setminus\; B\; =\; \backslash coprod\_\{i=1\}^n\; K\_i$ (relative complements can be written as finite disjoint unions).

The first property can be replaced with $\backslash mathcal\{S\}\; \backslash neq\; \backslash varnothing$ since $A\; \backslash in\; \backslash mathcal\{S\}\; \backslash implies\; A\; \backslash setminus\; A\; =\; \backslash varnothing\; \backslash in\; \backslash mathcal\{S\}.$

With the same notation, we call a family $\backslash mathcal\{R\}$ of subsets of $\backslash Omega$ a Ring of sets if it has the following properties:

• $\backslash varnothing\; \backslash in\; \backslash mathcal\{R\}$
• For all $A,\; B\; \backslash in\; \backslash mathcal\{R\},$ we have $A\; \backslash cup\; B\; \backslash in\; \backslash mathcal\{R\}$ (closed under pairwise unions)
• For all $A,\; B\; \backslash in\; \backslash mathcal\{R\},$ we have $A\; \backslash setminus\; B\; \backslash in\; \backslash mathcal\{R\}$ (closed under relative complements).

Thus, any ring on $\backslash Omega$ is also a semi-ring.

Sometimes, the following constraint is added in the measure theory context:

• $\backslash Omega$ is the disjoint union of a countable family of sets in $\backslash mathcal\{S\}.$

A field of sets (respectively, a semi-field) is a ring (respectively, a semi-ring) that also contains $\backslash Omega$ as one of its elements.

• Arbitrary (possibly uncountable) intersections of rings on $\backslash Omega$ are still rings on $\backslash Omega.$
• If $A$ is a non-empty subset of the powerset $\backslash mathcal\{P\}(\backslash Omega)$ of $\backslash Omega,$ then we define the ring generated by $A$ (noted $R(A)$) as the intersection of all rings containing $A.$ It is straightforward to see that the ring generated by $A$ is the smallest ring containing $A.$
• For a semi-ring $S,$ the set of all finite unions of sets in $S$ is the ring generated by $S:$ $$R(S)\; =\; \backslash left\backslash \{A\; :\; A\; =\; \backslash bigcup\_\{i=1\}^n\; A\_i,\; A\_i\; \backslash in\; S\backslash right\backslash \}$$ (One can show that $R(S)$ is equal to the set of all finite disjoint unions of sets in $S$).
• A content $\backslash mu$ defined on a semi-ring $S$ can be extended on the ring generated by $S.$ Such an extension is unique. The extended content can be written: $$\backslash mu(A)\; =\; \backslash sum\_\{i=1\}^n\; \backslash mu(A\_i)$$ for $A\; =\; \backslash bigcup\_\{i=1\}^n\; A\_i,$ with the $A\_i\; \backslash in\; S$ disjoint.

In addition, it can be proved that $\backslash mu$ is a pre-measure if and only if the extended content is also a pre-measure, and that any pre-measure on $R(S)$ that extends the pre-measure on $S$ is necessarily of this form.

In measure theory, we are not interested in semi-rings and rings themselves, but rather in σ-algebras generated by them. The idea is that it is possible to build a pre-measure on a semi-ring $S$ (for example Stieltjes measures), which can then be extended to a pre-measure on $R(S),$ which can finally be extended to a measure on a σ-algebra through Caratheodory's extension theorem. As σ-algebras generated by semi-rings and rings are the same, the difference does not really matter (in the measure theory context at least). Actually, Carathéodory's extension theorem can be slightly generalized by replacing ring by semi-field.{{cite book|last1=Klenke|first1=Achim|title=Probability Theory|date=2014|isbn=978-1-4471-5360-3|page=Theorem 1.53|doi=10.1007/978-1-4471-5361-0|series=Universitext}}

The definition of semi-ring may seem a bit convoluted, but the following example shows why it is useful (moreover it allows us to give an explicit representation of the smallest ring containing some semi-ring).

Think about the subset of $\backslash mathcal\{P\}(\backslash R)$ defined by the set of all half-open intervals $[a,\; b)$ for a and b reals. This is a semi-ring, but not a ring. Stieltjes measures are defined on intervals; the countable additivity on the semi-ring is not too difficult to prove because we only consider countable unions of intervals which are intervals themselves. Proving it for arbitrary countable unions of intervals is accomplished using Caratheodory's theorem.

Let $R$ be a ring of sets on $X$ and let $\backslash mu\; :\; R\; \backslash to\; [0,\; +\backslash infty]$ be a pre-measure on $R,$ meaning that $\backslash mu(\backslash varnothing)\; =\; 0$ and for all sets $A\; \backslash in\; R$ for which there exists a countable decomposition $A\; =\; \backslash coprod\_\{i=1\}^\backslash infty\; A\_i$ as a union of disjoint sets $A\_1,\; A\_2,\; \backslash ldots\; \backslash in\; R,$ we have

$$\backslash mu(A)\; =\; \backslash sum\_\{i=1\}^\backslash infty\; \backslash mu(A\_i).$$

Let $\backslash sigma(R)$ be the $\backslash sigma$-algebra generated by $R.$ The pre-measure condition is a necessary condition for $\backslash mu$ to be the restriction to $R$ of a measure on $\backslash sigma(R).$ The Carathéodory's extension theorem states that it is also sufficient,{{cite web|first=Noel|last=Vaillant|url=http://www.probability.net/WEBcaratheodory.pdf#page=16|title=Caratheodory's Extension|work=Probability.net|at=Theorem 4}} that is, there exists a measure $\backslash mu^\backslash prime\; :\; \backslash sigma(R)\; \backslash to\; [0,\; +\backslash infty]$ such that $\backslash mu^\backslash prime$ is an extension of $\backslash mu;$ that is, $\backslash mu^\backslash prime\backslash big\backslash vert\_R\; =\; \backslash mu.$ Moreover, if $\backslash mu$ is $\backslash sigma$-finite then the extension $\backslash mu^\backslash prime$ is unique (and also $\backslash sigma$-finite).{{cite book|first=Robert B.|last=Ash|title=Probability and Measure Theory|publisher=Academic Press|edition=2nd|year=1999|isbn=0-12-065202-1|page=19}}

First extend $\backslash mu$ to an outer measure $\backslash mu^*$ on the power set $2^X$ of $X$ by

$$\backslash mu^*(T)\; =\; \backslash inf\; \backslash left\backslash \{\backslash sum\_n\; \backslash mu\backslash left(S\_n\backslash right)\; :\; T\; \backslash subseteq\; \backslash cup\_n\; S\_n\; \backslash text\{\; with\; \}\; S\_1,\; S\_2,\; \backslash ldots\; \backslash in\; R\backslash right\backslash \}$$

and then restrict it to the set $\backslash mathcal\{B\}$ of $\backslash mu^*$-measurable sets (that is, Carathéodory-measurable sets), which is the set of all $M\; \backslash subseteq\; X$ such that $\backslash mu^*(S)\; =\; \backslash mu^*(S\; \backslash cap\; M)\; +\; \backslash mu^*(S\; \backslash cap\; M^\{\backslash mathrm\{c\}\})$ for every $S\; \backslash subseteq\; X.$

$\backslash mathcal\{B\}$ is a $\backslash sigma$-algebra, and $\backslash mu^*$ is $\backslash sigma$-additive on it, by the Caratheodory lemma.

It remains to check that $\backslash mathcal\{B\}$ contains $R.$ That is, to verify that every set in $R$ is $\backslash mu^*$-measurable. This is done by basic measure theory techniques of dividing and adding up sets.

For uniqueness, take any other extension $\backslash nu$ so it remains to show that $\backslash nu\; =\; \backslash mu^*.$ By $\backslash sigma$-additivity, uniqueness can be reduced to the case where $\backslash mu(X)$ is finite, which will now be assumed.

Now we could concretely prove $\backslash nu\; =\; \backslash mu^*$ on $\backslash sigma(R)$ by using the Borel hierarchy of $R,$ and since $\backslash nu\; =\; \backslash mu^*$ at the base level, we can use well-ordered induction to reach the level of $\backslash omega\_1,$ the level of $\backslash sigma(R).$

There can be more than one extension of a pre-measure to the generated σ-algebra, if the pre-measure is not $\backslash sigma$-finite, even if the extensions themselves are $\backslash sigma$-finite (see example "Via rationals" below).

Take the algebra generated by all half-open intervals [a,b) on the real line, and give such intervals measure infinity if they are non-empty. The Carathéodory extension gives all non-empty sets measure infinity. Another extension is given by the counting measure.

This example is a more detailed variation of the above. The rational closed-open interval is any subset of $\backslash mathbb\{Q\}$ of the form $[a,b)$, where $a,\; b\; \backslash in\; \backslash mathbb\{Q\}$.

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

Let $\backslash mu\_0$ be the counting set function ($\backslash \#$) defined in $\backslash Sigma\_0$.

It is clear that $\backslash mu\_0$ is finitely additive and $\backslash sigma$-additive in $\backslash Sigma\_0$. Since every non-empty set in $\backslash Sigma\_0$ is infinite, then, for every non-empty set $A\backslash in\backslash Sigma\_0$, $\backslash mu\_0(A)=+\backslash infty$

Now, let $\backslash Sigma$ be the $\backslash sigma$-algebra generated by $\backslash Sigma\_0$. It is easy to see that $\backslash Sigma$ is the $\backslash sigma$-algebra of all subsets of $X$, and both $\backslash \#$ and $2\backslash \#$ are measures defined on $\backslash Sigma$ and both are extensions of $\backslash mu\_0$. Note that, in this case, the two extensions are $\backslash sigma$-finite, because $X$ is countable.

Another example is closely related to the failure of some forms of Fubini's theorem for spaces that are not σ-finite.

Suppose that $X$ is the unit interval with Lebesgue measure and $Y$ is the unit interval with the discrete counting measure. Let the ring $R$ be generated by products $A\backslash times\; B$ where $A$ is Lebesgue measurable and $B$ is any subset, and give this set the measure $\backslash mu(A)\backslash text\{card\}(B)$. This has a very large number of different extensions to a measure; for example:

• The measure of a subset is the sum of the measures of its horizontal sections. This is the smallest possible extension. Here the diagonal has measure 0.
• The measure of a subset is $\backslash int\_0^1n(x)dx$ where $n(x)$ is the number of points of the subset with given $x$-coordinate. The diagonal has measure 1.
• The Carathéodory extension, which is the largest possible extension. Any subset of finite measure is contained in some union of a countable number of horizontal lines. In particular the diagonal has measure infinity.

• Outer measure: the proof of Carathéodory's extension theorem is based upon the outer measure concept.
• Loeb measures, constructed using Carathéodory's extension theorem.

{{reflist}}

{{PlanetMath attribution|id=5409|title=Hahn–Kolmogorov theorem}}

{{Measure theory}}

{{DEFAULTSORT:Caratheodory's Extension Theorem}}

Category:Theorems in measure theory