From: hook@nas.nasa.gov (Ed Hook)
Subject: Re: Sigma-rings
Date: 28 Sep 2000 22:57:56 GMT
Newsgroups: sci.math
Summary: [missing]

In article <39D39F67.F6BCE32C@nat.fr>,
            "Fabrice P. Laussy" <laussy@nat.fr> writes:
|> This is about Halmos' "Measure Theory", page 24, bottom of page,
|> Theorem D.

|> I understand all of the proof (which is 3 lines long!) but the
|> following argument (which is 2 lines long!!):

|> The union of all sigma-subrings of S which are generated by some
|> countable subclass of E is a sigma-ring.

|> (Notations: E is any class of sets, and S is the sigma-ring generated
|> by E).

|> When I say I don't understand it, it means I'm unable to prove it. I
|> do understand what it means, of course. It's certainly so trivial that
|> no proof is necessary but that isn't that obvious to me. A proof using
|> the definition is out of question (one is not to consider a
|> denumerable sequence of sets, each of which can be in E, or can be a
|> denumerable union of sets of E, or a difference of sets of E, or a
|> denumerable union of sets which are denumerable unions of sets of E,
|> or... any one of the infinite other combinations).

  Well, what you have to do is to verify
  the conditions that define a sigma-ring.
  But you don't need to explicitly characterize
  the detailed structure of the sets involved
  (as you seem to be fearing up above :-)
  For instance, the most difficult to verify
  is (probably) the assertion that the collection
  in question is closed under countable unions.
  So suppose that { A_n | n in N } is a countable
  subset of the alleged sigma-ring. By definition,
  each A_n lies in the sigma-subring generated
  by some countable collection E_n -- "clearly",
  then, all of the A_n lie in the sigma-subring
  generated by U E_k, so their union is an element
  of that sigma-subring. But U E_k is a _countable_
  subclass of E, so this sigma-subring is one of
  those that we're throwing into the mixture. That
  proves closure under countable unions ...

|> 
|> I guess we don't have such a result as the union of two (sigma)-ring
|> is a (sigma)-ring. So there is something to the fact it's about *all*
|> the unions. Intuitively, as a sigma-ring is made of denumerable
|> unions, taking the unions of denumerable class is only splitting the
|> matter. I cannot attach a formal meaning to it however. Can somebody
|> please help?

  Did that help ?

-- 
 Ed Hook                              |       Copula eam, se non posit
 Computer Sciences Corporation        |         acceptera jocularum.
 NAS, NASA Ames Research Center       |    All opinions herein expressed are
 Internet: hook@nas.nasa.gov          |              mine alone