Points of non-measurability
Dave L. Renfro
across in an old paper by R. L. Jeffery (in the middle
of a proof, and so it was pure chance that I noticed it)
that some of you might find interesting.
Given an arbitrary subset E of R^n, we say that x in R^n
is a point of non-measurability of E if every neighborhood
of x has a non-measurable intersection with E.
Is the set of points of non-measurability of a set E
(a) always, (b) sometimes, or (c) never
measurable?
[Hint: This is not very difficult.]
Dave L. Renfro
Robin Chapman
Isn't the set of points of non-measurability of E closed?
--
Robin Chapman
http://www.maths.ex.ac.uk/~rjc/rjc.html
"`Well, I'd already done a PhD in X-Files Theory at UCLA, ...'"
Greg Egan, _Teranesia_
Sent via Deja.com http://www.deja.com/
Before you buy.
bka...@cs.vu.nl
: In article <u5p6j0...@forum.swarthmore.edu>,
: dlre...@gateway.net (Dave L. Renfro) wrote:
:> Here's a problem based on a result I recently came
:> across in an old paper by R. L. Jeffery (in the middle
:> of a proof, and so it was pure chance that I noticed it)
:> that some of you might find interesting.
:>
:> Given an arbitrary subset E of R^n, we say that x in R^n
:> is a point of non-measurability of E if every neighborhood
:> of x has a non-measurable intersection with E.
:>
:> Is the set of points of non-measurability of a set E
:>
:> (a) always, (b) sometimes, or (c) never
:>
:> measurable?
:>
:> [Hint: This is not very difficult.]
: Isn't the set of points of non-measurability of E closed?
Certainly, if a point has a neigborhood whose intersection with
E is measurable, then all points in this neighborhood have a
neigborhood whose intersection with E is measurable (i.e. this
same neighborhood).
BK
: --
David C. Ullrich
Robin Chapman wrote:
> In article <u5p6j0...@forum.swarthmore.edu>,
> dlre...@gateway.net (Dave L. Renfro) wrote:
> > Here's a problem based on a result I recently came
> > across in an old paper by R. L. Jeffery (in the middle
> > of a proof, and so it was pure chance that I noticed it)
> > that some of you might find interesting.
> >
> > Given an arbitrary subset E of R^n, we say that x in R^n
> > is a point of non-measurability of E if every neighborhood
> > of x has a non-measurable intersection with E.
> >
> > Is the set of points of non-measurability of a set E
> >
> > (a) always, (b) sometimes, or (c) never
> >
> > measurable?
> >
> > [Hint: This is not very difficult.]
>
> Isn't the set of points of non-measurability of E closed?
I can prove the complement is open - maybe if I knew some set
theory I could see how to proceed from there.
Dave L. Renfro
[sci.math 25 Nov 99 16:14:11 -0500 (EST)]
<http://mathforum.com/epigone/sci.math/nenvochoi>
wrote (in part)
>Given an arbitrary subset E of R^n, we say that x in R^n
>is a point of non-measurability of E if every neighborhood
>of x has a non-measurable intersection with E.
>
>Is the set of points of non-measurability of a set E
>
>(a) always, (b) sometimes, or (c) never
>
>measurable?
As Robin Chapman, bkaster, and David C. Ullrich have
pointed out, the set is always closed. [Indeed, if the
set is measurable, you'll just have the empty set.]
It's the same type of argument that shows the set of limit
points, the set of condensation points, the set of points
of non-first category'ness, etc. are closed. Kuratowski
unifies this in his "Topology" treatise using the notion
of a local property.
It seems to me that this might make for a nice question
on a Ph.D. qualifying exam. It's not hard, but I bet
very few weak students will get it.
Here's a follow-up question, again not all that hard.
But it might be useful for anyone currently taking graduate
level real analysis course to think about it.
Find a set E contained in the reals having outer
measure less than delta (for any preassigned delta > 0)
such that the set of points of non-measurability of E
is the entire real line. [Note that you can't do better
than the empty set when delta = 0.] In fact, instead
of "outer measure < delta", we can require "Hausdorff
h-measure < delta for any preassigned Hausdorff measure
function h".
The first question I asked was simply mentioned
in passing by R. L. Jeffery (reference below), whereas
the question above is a slightly stronger version of a result
he proves in the middle of a proof of something else.
For those that might be interested, the paper by R.L.
Jeffery (also alluded to in my earlier post in this
thread) is
R. L. Jeffery, "Relative summability", Annals of Math.
(2) 33 (1932), 443-459. [See p. 459]
Dave L. Renfro
Dave L. Renfro
[28 Nov 99 19:48:28 -0500 (EST)]
<http://mathforum.com/epigone/sci.math/nenvochoi>
wrote (the foolish part)
>Find a set E contained in the reals having outer
>measure less than delta (for any preassigned delta > 0)
>such that the set of points of non-measurability of E
>is the entire real line. [Note that you can't do better
>than the empty set when delta = 0.]
So far, so good . . .
>In fact, instead of "outer measure < delta", we can
>require "Hausdorff h-measure < delta for any preassigned
>Hausdorff measure function h".
Ouch!! Pretend you didn't read this last sentence.
No "fractal dimension" improvement is possible.
Once you're measure zero, that's it--the set of
non-measurability is the empty set.
Dave L. Renfro