2
$\begingroup$

Wikipedia's article(https://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_extension_theorem) says:

Let $R$ be a ring on $\Omega$ and $\mu: R \rightarrow [0, +\infty]$ be a pre-measure on $R$. The Carathéodory's extension theorem states that[2] there exists a measure $\mu':\sigma (R) \rightarrow [0, +\infty]$ such that $\mu'$ is an extension of $\mu$. (That is, $\mu'|R = \mu$). Here $\sigma (R)$ is the $\sigma$-algebra generated by $R$. If $\mu$ is $\sigma$-finite then the extension $\mu'$ is unique (and also σ-finite).[3]

Let $(X,\mathcal A,\nu)$ be a finite measure space, where $\mathcal A$ is a $\sigma$-algebra. Let $Y$ be a non-empty member of $\mathcal A$ such that $X\neq Y$. Let $\mathcal R=\{E:E\subset Y,E\in \mathcal A\}$. Then $\mathcal R$ is a $\sigma$-ring on $X$. Let $\mu$ be the restriction of $\nu$ on $\mathcal R$. Let $\mathcal B = \{E: E\in \mathcal R \text{ or } X- E\in \mathcal R\}$. $\mathcal B$ is the $\sigma$-algebra generated by $\mathcal R$. Since $\mathcal R \subset \mathcal A$, $\mathcal B \subset \mathcal A$. Let $\nu'$ be the restriction of $\nu$ on $\mathcal B$. Then $\mu$ is a measure on $\mathcal R$ and $\nu'$ is an extension of $\mu$.

Define $\lambda: \mathcal B \rightarrow [0, +\infty]$ as follows. If $E\in \mathcal R$, then define $\lambda(E) = \mu(E)$, otherwise define $\lambda(E) = \infty$. Then $\lambda$ is a measure on $\mathcal B$ and it is an extension of $\mu$. Since $\nu'(X) \lt \infty$ and $\lambda(X) = \infty$, $\nu' \neq \lambda$. This seems to be a counter-example of Wikipedia's theorem above.

Am I mistaken?

$\endgroup$
  • $\begingroup$ You $\mu$ (i.e. $\nu |_{\mathcal{R}}$) is not $\sigma$ finite. You can not write $X = \bigcup_n E_n$ with $E_n \in \mathcal{R}$, since any $E \in \mathcal{R}$ satisfies $E \subset Y \subsetneq X$. $\endgroup$ – PhoemueX Jul 25 '15 at 21:38
  • $\begingroup$ @PhoemueX Your definition of $\sigma$-finiteness of a measure is commonly accepted when a measure is defined on an algebra or a $\sigma$-algebra. However, if a measure is defined on a ring or $\sigma$-ring, some authors(for example Halmos) define $\sigma$-finiteness as follows. When a measure $\mu$ is defined on a ring $R$ on a set $X$, $\mu$ is called a finite measure if $\mu(E)\lt \infty$ for all $E\in R$. It is said to be $\sigma$-finite if every member of $R$ is a countable union of members of finite measure. $\endgroup$ – JHW Jul 25 '15 at 22:23
  • $\begingroup$ Ok, I did not know that. But at least the source cited in the Wikipedia article (Ash) uses the definition of sigma-finiteness that I mentioned above :) $\endgroup$ – PhoemueX Jul 26 '15 at 5:53
  • $\begingroup$ @PhoemueX Ash only defines $\sigma$-finiteness of measures that are defined on algebras. He does not mention $\sigma$-finiteness of measures defined on rings, does he? $\endgroup$ – JHW Jul 26 '15 at 7:03
  • $\begingroup$ Yes, but that makes the whole Wiki article even more doubtable. They cite Ash for the uniqueness regarding sigma finite measures on rings, while he only coniders 'fields'. I just wanted to mention that on page 19 (which Wiki references) he explicitly gives the definition I mentioned above. $\endgroup$ – PhoemueX Jul 26 '15 at 7:04
1
$\begingroup$

Seems to me you're right. The version of the theorem I know talks about algebras instead of rings; then your counterexample doesn't work.

Hmm. Of course your $\lambda$ is also typically not going to be $\sigma$-finite. Seems to me another way to fix the theorem is to say that if $\mu$ is $\sigma$-finite then there is a unique $\sigma$-finite extension.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.