1
$\begingroup$

The Caratheodory method provides the following relationship between measures and outer measures. For the definitions, I am following [Fremlin, "Measure Theory"].

Let $X$ be a set. Denote by $\mathbf{M}_X$ the collection of positive measures $\mu : \Sigma \to [0, \infty]$ defined on a $\sigma$-algebra $\Sigma \subseteq \mathcal{P}(X)$ and by $\mathbf{O}_X$ the collection of outer measures $\eta : \mathcal{P}(X) \to [0, \infty]$ (i.e. $\eta(\emptyset) = 0$, $\eta$ is non-decreasing and countably subadditive). Let us denote the elements $\mu : \Sigma \to [0, \infty]$ in $\mathbf{M}_X$ by $(\Sigma, \mu)$. The Caratheodory method provides the mappings

$F : \mathbf{M}_X \to \mathbf{O}_X$, $(\Sigma, \mu) \mapsto \mu^*$, $\mu^*(E) := \inf \{ \sum_{n=1}^\infty \mu(A_n) \mid E \subseteq \bigcup_{n=1}^\infty A_n, A_n \in \Sigma \}$ (the induced outer measure) and

$G : \mathbf{O}_X \to \mathbf{M}_X$, $\eta \mapsto (M(\eta), \eta|_{M(\eta)})$ where $M(\eta) := \{ E \in \mathcal{P}(X) \mid \eta(A) = \eta(A \cap E) + \eta(A \cap E^c) \textrm{ for all } A \in \mathcal{P}(X) \}$ is the $\sigma$-algebra of Caratheodory $\eta$-measurable sets.

A measure $(\Sigma, \mu)$ is complete if $\Sigma$ contains all subsets of all $\mu$-null sets. An outer measure $\eta$ is regular if $FG(\eta) = \eta$, i.e. the outer measure induced from $\eta|_{M(\eta)}$ by the Caratheodory construction is the initially given outer measure $\eta$ [Fremlin, 132Xa].

Facts:

  1. $FG(\eta) \geq \eta$ for all outer measures $\eta$. In general, $FG \neq id$.
  2. Every induced outer measure $\mu^* = F(\Sigma, \mu)$ is regular and conversely, every regular outer measure $\eta$ is induced by some measure $(\Sigma, \mu)$, i.e. $\eta = F(\Sigma, \mu)$, just take $(\Sigma, \mu) := G(\eta)$ [Fremlin, 132Xa]. Hence, the range of $F$ is precisely the collection of regular outer measures.
  3. It holds $FGF = F$.
  4. Every induced measure $(\Sigma, \mu) = G(\eta)$ is complete, hence the range of $G$ is contained in the collection of complete measures. In contrast, a complete measure $(\Sigma, \mu)$ need not be induced by some outer measure. However, $(\check{\Sigma}, \check{\mu}) := GF(\Sigma, \mu)$ is an extension of $(\Sigma, \mu)$ to a complete measure, i.e. $\Sigma \subseteq \check{\Sigma}$ and $\check{\mu} = \mu$ on $\Sigma$ for any measure $(\Sigma, \mu)$ (complete or not).
  5. In general, $GFG \neq G$.

[Fremlin, 213Ya] provides an example for 5.: Let $X = \mathbb{N}$ and define $\eta$ by $\eta(A) := \sqrt{|A|}$ if $A$ is finite and $\eta(A) := \infty$ if $A$ is infinite. Then $\eta$ is an outer measure, that is not regular; for the induced measure $(\Sigma, \mu) := G(\eta)$ it holds $\Sigma = \{ \emptyset, X \}$; its induced outer measure $\mu^* := G(\Sigma, \mu) = FG(\eta)$ is given by $\mu^*(E) = \infty$ for any $E \neq \emptyset$ and in turn $GFG(\eta) = (\mathcal{P}(X), \eta) \neq (\Sigma, \mu) = G(\eta)$.

However, [Fremlin, 213Xa(v)] states that "if $\mu$ is defined by Carathéodory's method from another outer measure, then $\mu = \check{\mu}$"; in other words $GFG = G$. So, Fremlin contradicts himself here. $GFG = G$ holds at least for regular outer measures.

The equations $FGF = F$ and "$GFG = G$" indicate that there should be some hidden Galois connection between a subset $\mathbf{M}_X' \subseteq \mathbf{M}_X$ and a subset $\mathbf{O}_X' \subseteq \mathbf{O}_X$ defined by the Caratheodory method, so that certain complete measures and regular outer measures form the fixed points of the Galois connection (and are equivalent in this way). An answer to this thread seems to go in this direction. The Galois connection should of course be not trivial in the sense that it only consists of the fixed points. Moreover, we need particular partial orders on $\mathbf{M}_X'$ and $\mathbf{O}_X'$. Does anyone has an idea how to do that?

$\endgroup$
1
  • $\begingroup$ I agree that 213 Ya (in the current version: www1.essex.ac.uk/maths/people/fremlin/chap21.pdf 213Yb) contradicts 213X(a) (v), and that 213Y (a) is the correct one. If you contact him about it he can probably correct it in the online version. $\endgroup$ – Robert Furber Oct 16 '19 at 15:00

Your Answer

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

Browse other questions tagged or ask your own question.