Is a sigma ring somehow the closure of a ring?

[sto]

Let X be a set and R_ be a nonempty class of subsets of X having the
properties:

E in R_ and F in R_ => E\F in R_
E in R_ and F in R_ => union(E,F) in R_

Call R_ a ring (of subsets of X) and denote by S_ the sigma ring
generated by R_.

Let m be a measure on S_. There is a theorem that states that given any
e>0 and E in S_, there exist some E0 in R_ such that

m ( union(E\F,F\E) ) < e

My question is can we conclude from this that the sets in R_ are somehow
dense in S_?

If we form an equivalence class ~ by defining

E~F iff m( union(E\F,F\E) ) = 0

and then considering m to be a function defined on this equivalence
class, it seems to me that we turn (S_,m) into a metric space and that
in this metric space every open set in S_ intersects R_. But is this
technically correct? And if it is not, is it nonetheless still a good
way to think about the "topology" (if there is such a thing) of S_ and R_?

THanks,
-sto

[William Elliot]

On Tue, 28 Dec 2010, sto wrote:

> Let X be a set and R_ be a nonempty class of subsets of X having the
> properties:
>
> E in R_ and F in R_ => E\F in R_
> E in R_ and F in R_ => union(E,F) in R_

> Call R_ a ring (of subsets of X) and denote by S_ the sigma ring
> generated by R_.

> Let m be a measure on S_. There is a theorem that states that given any e>0
> and E in S_, there exist some E0 in R_ such that
>
> m ( union(E\F,F\E) ) < e
>

The theorem as stated makes no sense.

[sto]

>> Let X be a set and R_ be a nonempty class of subsets of X having the
>> properties:
>>
>> E in R_ and F in R_ => E\F in R_
>> E in R_ and F in R_ => union(E,F) in R_
>
>> Call R_ a ring (of subsets of X) and denote by S_ the sigma ring
>> generated by R_.

>> Let m be a measure on S_. There is a theorem that states that given
>> any e>0 and E in S_, there exist some E0 in R_ such that
>>
>> m ( union(E\F,F\E) ) < e
>>
> The theorem as stated makes no sense.
>

Why not? Is it because

m ( union(E\F,F\E) ) < e

should read

m ( union(E\E0,E0\E) ) < e

or is there another reason? (Also, union(E\E0,E0\E) just denotes the
symmetric difference between E and E0. I can't think of any other way
to write that.)

I looked up the theorem and it turns out that m has to be sigma-finite
and E has to be of finite measure: "If m is a sigma-finite measure on a
ring R_, then, for every set E of finite measure in S_ and for every
positive number e, there exists a set E0 in R_ such that

m( union(E\E0,E0\E) ) <= e "
Halmos, *Measure Theory* p.56 This is not exactly the same statement as
in my post because of the conditions on m and E, but nonetheless it
still looks to me like it says that every "open set" in S_ "intersects"
R_ suggesting that there exists some sense in which S_ could be the
closure of R_.

[William Elliot]

On Wed, 29 Dec 2010, sto wrote:

>>> Let X be a set and R_ be a nonempty class of subsets of X having the
>>> properties:
>>>
>>> E in R_ and F in R_ => E\F in R_
>>> E in R_ and F in R_ => union(E,F) in R_
>>
>>> Call R_ a ring (of subsets of X) and denote by S_ the sigma ring
>>> generated by R_.
>
>>> Let m be a measure on S_. There is a theorem that states that given
>>> any e>0 and E in S_, there exist some E0 in R_ such that
>>>
>>> m ( union(E\F,F\E) ) < e
>>>
>> The theorem as stated makes no sense.
> Why not? Is it because
>
> m ( union(E\F,F\E) ) < e
> should read
> m ( union(E\E0,E0\E) ) < e
>

Yes, in part.

> or is there another reason? (Also, union(E\E0,E0\E) just denotes the
> symmetric difference between E and E0. I can't think of any other way to
> write that.)
>
> I looked up the theorem and it turns out that m has to be sigma-finite and E
> has to be of finite measure: "If m is a sigma-finite measure on a ring R_,
> then, for every set E of finite measure in S_ and for every positive number
> e, there exists a set E0 in R_ such that

> m( union(E\E0,E0\E) ) <= e "

Indeed, there needs to be some conditions on m.

Let R be an algebra, S the sigma ring generated by R
and m a sigma-finite measure on S. Then for all E in S
of finite measure and every e > 0, there's some D in R with

m(E\D \/ D\E) < e.

Is that how the theorem is suppose to read?

> Halmos, *Measure Theory* p.56 This is not exactly the same statement as in my
> post because of the conditions on m and E, but nonetheless it still looks to
> me like it says that every "open set" in S_ "intersects" R_ suggesting that
> there exists some sense in which S_ could be the closure of R_.
>

I'll not even attempt to understand your
talking out loud, stream of thought.

>>> My question is can we conclude from this that the sets in R_ are
>>> somehow dense in S_?

Dense in what sense?

>>> If we form an equivalence class ~ by defining
>>>
>>> E~F iff m( union(E\F,F\E) ) = 0

D\E \/ E\D
E\F \/ F\E

D\E /\ E\F .\/. E\D /\ E\F .\/. D\E /\ F\E .\/. E\D /\ F\E
E\D /\ E\F .\/. D\E /\ F\E

Why is ~ transitive?
Is ~ a relation for R or for S?

>>> and then considering m to be a function defined on this equivalence
>>> class, it seems to me that we turn (S_,m) into a metric space and that
>>> in this metric space every open set in S_ intersects R_.

What's the metric? d(E/~, F/~) = m(E\F \/ F\E)?

>>> But is this technically correct?

I don't know. What have you done to show d is a metric?
In particular, why the triangle inequality?

>>>>And if it is not, is it nonetheless still a good

>>> way to think about the "topology" of S_ and R_?

Where there's a metric, there's a topology.
Wouldn't it be the topology of S/~ or of R/~?

----