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?

  • $\begingroup$ 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

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.


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