Talk:Carathéodory's extension theorem

I added a small stub on Caratheodoy's extension theorem. This is a fundamental theorem, but unfortunately, I don't know much about topology and measure theory outside the probabilistic context; I am sure there are some links with those fields I cannot do myself. Ashigabou 03:27, 11 February 2006 (UTC)

In measure theory, Carathéodory's extension theorem proves that

for a given set Ω, you can always extend a measure defined on R

to the σ-algebra generated by R, where R is a ring included in

the power set of Ω;

In contrast to the extension of a measure on a semi-ring to a measure on a ring, the above requires that there exist a monotonically increasing sequence $(A\_n)\_\{n\; \backslash in\; \backslash mathbb\{N\}\}$, such that $\backslash Omega\; =\; \backslash bigcup\_\{n\; =\; 1\}^\backslash infty\; A\_n$. Every subset of $\backslash Omega$ can then be covered by a countable union of sets, which is required to define the outer measure that will subsequently be reduced to the desired measure on the generated $\backslash sigma$-Algebra.

moreover, the extension is unique.

The extension to a $\backslash sigma$-Algebra is unique only if the measure on the generating ring is $\backslash sigma$-finite. Note that this already implies the condition above. I therefore suggest to simply add the $\backslash sigma$-finiteness to the conditions of the theorem. --Drizzd 10:35, 8 February 2007 (UTC)

: Since there were no objections I added the suggested condition to the article. --Drizzd 12:31, 19 February 2007 (UTC)

"Warning: I've seen the following theorem called the Carathéodory

extension theorem, the Carathéodory-Fréchet extension theorem,

the Carathéodory-Hopf extension theorem, the Hopf extension

theorem, the Hahn-Kolmogorov extension theorem, and many others that

I can't remember! We shall simply call it Extension

Theorem. However, I read in Folland's book (p. 41) that the theorem

is originally due to Maurice René Fréchet (1878–1973) who

proved it in 1924."

[http://www.math.binghamton.edu/paul/505-S08/505-7.pdf Paul Loya] (page 33).

Boris Tsirelson (talk) 16:15, 9 November 2013 (UTC)

: The article on the Caratheodory and Hahn-Kolmogorov theorems should be merged. 129.215.104.198 (talk) 18:07, 9 October 2015 (UTC)

::I have just now merged from Hahn-Kolmogorov theorem just now. The talk page there listed four different people stating it should be merged, plus the statement above makes five, plus my own reading makes six so that is six votes to merge. For other comments, refer to Talk:Hahn-Kolmogorov theorem. In particular, there is some burblings as to which variants are more general than others: Is the statement using semi-rings more general, or less, than the one for field of sets? I can't quite tell, off-hand. 67.198.37.16 (talk) 02:28, 4 October 2020 (UTC)

It is claimed that

"If μ is σ-finite then the extension μ′ is unique (and also σ-finite)."

Unless I have made a mistake, the claim inside parentheses is false. Consider the counting measure μ defined on the ring R of finite subsets of the real line. It is finite (and hence σ-finite), but its extension μ′ is not, since the real line (which is in the σ-field generated by R) is not a countable union of finite sets.

The extension to the σ-ring generated by R is σ-finite, however (see, e.g., Theorem 2.5.3 from "A basic course in measure and probability: theory for applications", by Ross Leadbetter, Stamatis Cambanis and Vladas Pipiras). In particular, if the whole space is a countable union of elements from the ring, then μ′ is σ-finite.

Note that the last condition (actually, something apparently slightly stronger, since the union is required to be disjoint) is required under the alternative definition for rings given in the first section of this article, but it doesn't seem to be made explicit that such condition is needed if we want to conclude μ′ is σ-finite. — Preceding unsigned comment added by TuringMachine001 (talk • contribs) 15:44, 15 February 2016 (UTC)

The second measure presented is:

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.

However, since $Y$ is the unit interval with the discrete counting measure, it is not $\backslash sigma$-finite. So, we cannot conclude that the function $x\; \backslash mapsto\; n(x)$ is measurable. If it is not measurable, then $\backslash int\_0^1n(x)dx$ is not defined.

Example: Let $C=[0,1]\backslash times\; E$, where $E$ is any non-Lebesgue measurable subset. Clearly $C\; \backslash in\; R$ (because $R$ is the ring generated by products $A\backslash times\; B$ where $A$ is Lebesgue measurable and $B$ is any subset). Let $D\; =\; \backslash \{(x,x)\; :\; x\; \backslash in\; [0,1]\; \backslash \}$. It is easy to see that $D\; \backslash in\; \backslash sigma(R)$. Let $S\; =\; C\; \backslash cap\; D$. It follows immediately that $S\; \backslash in\; \backslash sigma(R)$,

$S\; =\; \backslash \{(x,x)\; :\; x\; \backslash in\; E\; \backslash \}$ and $n\_S$ is the indicator function of $E$ and so it is not a measurable function.

50.100.95.67 (talk) 05:28, 26 March 2023 (UTC)

I do not see why both "finite additivity" and "sigma additivity" are written there.

Sigma additivity should imply finite additivity, just replace all $\backslash infty$ with $N$. If you really need infinite sequences, then add as many empty sets as you need, as $\backslash empty$ is has to be even in a semialgebra, not just algebra.