Edwin Chan

Posted on Monday, 07 July, 2003 - 02:58 am:

Could someone give me examples of:

1. A countably infinite union of closed sets that is not closed

2. In probability theory in the continuous setting, there is the statement that not ALL subsets of a continuous space can be considered as EVENTS i.e. subsets for which a probability measure satisfying the standard criteria may be assigned, thereby requiring a very careful specification of the kinds of "admissable" subsets of the sample space, resulting in the sample space with the properties of a Borel field (sigma algebra/completely additive field). Can someone actually give me an example of a subset in a continuous space that cannot be assigned a probabilty measure, I suspect they must be quite weird creatures? I'd also like some insight on what actually is the "cause" of the problem - I've read somewhere that if you use the OUTER measure ALL subsets can be assigned probabilities but probabilities of mutually exclusive events are only sub-additive. So if you use the INNER measure (is this the same as the Lebesgue measure?) then you recover the additive property at the expense of not "covering" all subsets - so back to my request for an example of these elusive subsets.
Thanks very much.

David Loeffler

Posted on Monday, 07 July, 2003 - 04:21 am:

For (1), note that singletons are closed sets, so just take a sequence of single points tending to some other point, for example {1/n : n in

} in

with the usual metric.

Edwin Chan

Posted on Monday, 07 July, 2003 - 05:23 am:

Thanks David, that was a great help.

Edwin Chan

Posted on Monday, 07 July, 2003 - 08:16 am:

A supplementary question that has also been bugging me, how does one reason from the definition of a closed set:

If all limit points of E are points of E, then E is closed

to the conclusion that finite point sets are closed, when finite point sets have NO limit points for one to classify as being inside or outside of E?

Demetres Christofides

Posted on Monday, 07 July, 2003 - 08:30 am:

I think that to find such a set you have to use the axiom of choice. [I am not 100% sure about this statement. Perhaps somebody else can either confirm it or correct me.]

The simplest example I know is the following:

Consider the space $S^{1}$ (unit circle) with the measure of the subset $A$ of $S^{1}$ to be $m (A) =$ area of $A$ .

Define an equivalence relation $~$ on $S^{1}$ as following:

$e^{ix} ~ e^{iy}$ iff $x - y$ is rational.

This partitions $S^{1}$ into equivalence classes. Pick one element from each class to form a set $A$ (use axiom of choice).

Note that for any rational $q$ , $A + q = {a + q : a \in A}$ is disjoint from $A$ (OK $q \neq 0$ ) and also for distinct rational $p$ , $q$ $A + p$ and $A + q$ are disjoint. Show that if $m (A)$ exists then $m (A) = m (A + q)$ .

Show that $S^{1} = ⋃_{q} (A + q)$ where the union is disjoint and over all rationals $q$ .

But then (countable additivity) $2 π = m (S^{1}) = \sum_{q} m (A + q) = \sum_{q} m (A)$ , giving a contradiction.

Demetres

Edwin Chan

Posted on Monday, 07 July, 2003 - 09:48 am:

Thanks Demetres, am I to understand that this set A formed by using the axiom of choice is an example of a subset of a space that cannot be assigned a unique measure?

If so, could you explain the significance of/need for invoking the axiom of choice to form this set A?

What motivated you to define the equivalence relation as such and could you clarify why such a definition makes for an equivalence realtion?

David Loeffler

Posted on Monday, 07 July, 2003 - 10:34 am:

If there aren't any limit points, it is vacuously true that all limit points of E are in E. If A is always false, 'A Ã?? B' is true for any B.

David

Tim Austin

Posted on Monday, 07 July, 2003 - 01:17 pm:

Hi folks. Any subset of any set will be be measured by some measure - for example, for any non-empty set S and any point x in S you can consider the delta measure d_x defined for all subsets of S by

d_x (E) = 1 if x is in E
= 0 else.

It's easy to see that this is a measure on the family of all subsets of S. However, when we look at, say, the unit interval or S^1, and ask about continuity, we restrict ourselves to considering certain kinds of measure; in particular those which are said to be absolutely continuous with respect to Lebesgue measure (if m is Lebesgue measure and n is our other measure defined on some sigma-algebra which at least contains all the Lebesgue measurable sets, n is absolutely continuous with respect to Lebesgue measure if m(E)=0 implies n(E)=0 for any Lebesgue-measurable E. The Radon-Nikodym theorem says that any such measure n is of the form

Latex image click or follow link to see src

for some Lebesgue-integrable function f; in probabilistic language, n has a p.d.f. with respect to m).

The point about Lebesgue measure (on the real line, the unit interval with addition taken mod 1, or in the natural way on S^1) is that it is translation invariant, so any translate of a Lebesgue measurable set E will have the same Lebesgue measure as E. This is why Demetres' set A cannot be measurable for Lebesgue measure, as we have countably many copies of it all obtained by translation, with union some set with finite non-zero measure, so we hit the problem Demetres explained. I suppose this is the motivation for that example: you want to use the translation invariance (which can in fact be shown to characterize Lebesgue measure uniquely among all measures which measure all open sets, although I think that's a slightly deep fact) to get your counterexample.

Incidentally, it is true that you need the axiom of choice to construct a non-Lebesgue measurable set (although I don't know a proof of this). A related issue is the Banach-Ulam problem, one of the most important open questions in abstract measure theory:

Is there a set X and a probability measure m defined for /all/ subsets of X such that m(E) = 0 for any singleton E?

As far as I know, it is suspected but not known that the answer to this question is independent of all the standard axioms of set theory (including choice).

P.S. on a technical note, an INNER measure is usually taken to mean something different again from an outer measure or just a measure, involving countable superadditivity and some fiddly continuity conditions; they don't come up very often as they're not useful for all that much.

Tim