Let $X$ be a set.

Let $\mathcal{R}$ be a set of subsets of $X$ such that

$\{\} \in \mathcal{R}$


For all members $A$ and $B$ of $\mathcal{R}$, $\;\; (A\cup B)-(A\cap B) \; \in \; \mathcal{R} \;\;$.


For all sequences $\; \langle A_0,A_1,A_2,A_3,...\rangle \;$ of members of $\mathcal{R}$, $\;\;\;\; \displaystyle\bigcap_{n=0}^{\infty} \; A_n \;\; \in \;\; \mathcal{R}$


Let $\;\; \phi : \mathcal{R} \to \mathbb{R} \;\;$ be such that for all pairwise disjoint
sequences $\; \langle A_0,A_1,A_2,A_3,...\rangle \;$ of members of $\mathcal{R}$,

if $\;\;\;\; \displaystyle\bigcup_{n=0}^{\infty} \; A_n \;\; \in \;\; \mathcal{R} \;\;\;\;$ then $\;\;\;\; \displaystyle\sum_{n=0}^{\infty} \; \phi(A_n) \;\; = \;\; \phi\left(\displaystyle\bigcup_{n=0}^{\infty} \; A_n\right) \;\;\;\;$.

Does it follow that there exists a member $P$ of $\mathcal{R}$ such that
for all members $A$ of $\mathcal{R}$, $\;\; A\cap P \; \in \; \mathcal{R} \;\;$ and $\; \phi(A-P\hspace{.01 in}) \leq 0 \leq \phi(A\cap P\hspace{.01 in}) \;\;$?

  • 1
    $\begingroup$ I must be missing a subtlety here. The standard argument works on $\sigma$-rings. This is done e.g. in Halmos's measure theory, Section 29. $\endgroup$ – Theo Buehler Feb 29 '12 at 9:55
  • $\begingroup$ It seems to me that the definition here of $\sigma$-ring is not that given by Wikipedia: en.wikipedia.org/wiki/Sigma-ring (which agrees with that given by Halmos). For example, if $\mathcal R$ is the collection of all finite subsets of $X$, then that's allowed for definition in the OP, but not for Halmos... Indeed, you seem to have defined a Delta Ring: en.wikipedia.org/wiki/Delta_ring Wikipedia claims that these are enough for measure theory. But it does seem to me that Halmos's proof of the Hahn-Decomposition doesn't work (I could well be wrong...) $\endgroup$ – Matthew Daws Feb 29 '12 at 13:18

Your axioms are:

  1. $\emptyset\in\mathcal R$
  2. $A,B\in\mathcal R \implies (A\cup B)\setminus (A\cap B)\in\mathcal R$
  3. $A_n\in\mathcal R \implies \bigcap_n A_n\in\mathcal R$

Then 2 and 3 show that $A,B\in\mathcal R \implies A\setminus B\in\mathcal R$. Then 2 shows that $\mathcal R$ is closed under disjoint unions, so combined with 3, we see that $A,B\in\mathcal R \implies A\cup B = (A\setminus B) \cup (B\setminus A) \cup (A\cap B) \in \mathcal R$. So actually your axioms describe a "$\delta$-ring" (see http://en.wikipedia.org/wiki/Delta-ring ) and not a $\sigma$-ring.

Here then is a counter-example to your question. Let $X=\mathbb N$ with $\mathcal R$ being the collection of all finite subsets of $X$. Define $\phi$ by \[ \phi(A) = |A\cap\{\text{evens}\}| - |A\cap\{\text{odds}\}|. \] This is a "measure" in your sense. But then the only choice for $P$ would be the set of even numbers, and that's not in $\mathcal R$.

For a positive result, note that for any $A\in\mathcal R$, the collection $\mathcal R_A = \{ A\cap B : B\in\mathcal R \}$ is a $\sigma$-algebra on $A$, and $\phi$ restricted to $A$ is a signed measure in the usual sense. So there is a Hahn-Decomposition for $A$. You might think one could glue the Hahn decompositions together. Perhaps getting $P\subseteq X$ such that for all $B\in\mathcal R$, we did have $B\cap P, B\setminus P \in\mathcal R$ and $\phi(B\setminus P) \leq 0 \leq \phi(B\cap P)$. (This is true in my example). But my gut reaction suggests that this probably can't be done in general (I await a good counter-example from someone else!)


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