Re: Measure and Compactness

David C. Ullrich wrote:

On Tue, 11 Apr 2006 06:59:18 EDT, Maury Barbato
<mauriziobarbato@xxxxxxxx> wrote:

Hello,
I want to say first of all that I have scarce
knowledge
of functional analysis, so forgive me for my maybe
trivial question.
Let X be a bounded Jordan measurable subset of R^n
(I
remeber, since Jordan measure is not so known today,
that
a bounded subset of R^n is Jordan measurable, if its
boundary has Lebesgue measure zero). Let J be the
set
of all Jordna measurable subsets of X. For any x,y
in J,
we put
d(x,y)=measure of symmetric difference of x and y.
If we consider x, y in J equivalent if d(x,y)=0, we
obtain an equivalence relations. The set S of all
equivalent classes, with the distance above, is a
metric
space.
The questions are: is S compact? Is it complete?

It seems pretty clear that it's not complete and
hence
not compact, simply because, for example, any
Lebesgue
measurable subset of [0,1] can be approximated in
this metric by Jordan measurable sets.

Does it
have for any e>0 a finite e-net?

No. Consider X = [0,1]. Let E_n be the union
of the intervals [j/2^n, (j+1)/2^n] for
odd j. I think it's pretty clear that
d(E_n, E_m) = 1/2 for all n <> m.

What if we replace Jordan measure with Lesbegue
measure
everywhere?

Consider E_n as above.

Thank you very very much for your attention.
Maury


************************

David C. Ullrich

Thank you very very much for your invaluable help. I was
exploring these properties of S since with regard to
the following problem.

Let X be a bounded Jordan measurable subset of R^n
and f:X->R a Riemann integrable function. If the
integral of f over a Jordan measurable subset Y of X is
a and the integral of f over another Jordan measurable
subset Z of X is b, with a<b, does there exists, for
every a<c<b, a Jordan measurable subset C of X such that
the integral of f over C is c?
Once I was quite convinced that the answer is yes, but
now I have some doubts.

Thank you very very much for your attention.
My Best Regards,
Maury
.

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