When does a nowhere dense set have non-zero measure?
Peter
I know that nowhere dense sets can have positive measure (such as
Cantor sets). If I have a specific subset of a line segment which I
know is nowhere dense, how can I determine it it has positive measure?
Thank-you,
Peter
David C. Ullrich
wrote:
At this level of generality there's really nothing that can be said.
It has positive measure if the measure is positive.
If you have a more specific set that you're wondering about someone
may have a clue.
>Thank-you,
>
>Peter
Peter
> On Thu, 15 Apr 2010 19:37:11 -0700 (PDT), Peter <pwoly...@yahoo.com>
Hi David,
OK, so there's no general condition one can test.
It has to do with the set of points whose trajectories generated by a
1D discrete dynamical system do not converge. This set, which I'll
denote by X, is nowhere dense because the map has 3 properties:
1) It has a single basin of attraction
2) It maps the interval [-pi/2,pi/2] back into itself
3) Outside the basin of attraction the map is expansive
These conditions are sufficient to guarantee that no interval is
contained in X.
Unfortunately, I think the problem itself is probably too hard for
anyone to answer casually on this ng. However, if anyone has an actual
research interest in this type of problem though, they can contact me
via my e-mail address.
Thanks,
Peter
junoexpress
<pwoly...@yahoo.com>
> wrote:
>
> >Hi,
>
> >I know that nowhere dense sets can have positive measure (such as
> >Cantor sets). If I have a specific subset of a line segment which I
> >know is nowhere dense, how can I determine it it has positive measure?
>
Hi Peter
There may be some things you can say about your set of points X. For
example, do you know if the set is countable? If you can demonstrate,
for example, that the set is countable, then you can say it has
measure zero. The reason why I raise this issue is that the type of
thing it seems that you are describing involves a stretching, folding
and then removing a portion of the interval. Such a process may give
rise to a set having some properties of a Cantor set. I don't recall
the details, so this response may be off a bit, but I seem to recall
some type of general argument based on a ternary decimal system
regarding the middle third Cantor sets that showed the set of points
not removed by the constriction was not countable. I don't know the
details off hand, but this may be something you want to look at and
see if it fits your problem.
Good Luck,
Matt
Dave L. Renfro
If, in fact, the set happens to have zero measure, here's a commonly
used strategy that I outline in the following 20 February 2001
sci.math.research post.
-----------------------------------------------------------
-----------------------------------------------------------
http://groups.google.com/group/sci.math.research/msg/c7f8cef7ae8cfd11
Robert Valkenburg wrote:
> I wish to show a particular subset of a compact set in R^n
> has measure zero. I am fairly sure the subset is nowhere dense.
>
> Under what conditions do nowhere dense sets in R^n have
> measure zero?
A common method for showing that a set has measure zero makes use
of the Lebesgue density theorem. The Lebesgue density theorem
says that at almost every (i.e. all but a set of measure zero)
point x in a set E the following holds:
limit as R --> 0 of
[measure of E intersect B(x,R)] / [measure of B(x,R)]
equals 1.
Your set is probably measurable, so I doubt measurability is an
issue. However, if measurability is in doubt, the result still
holds with "outer measure" in place of "measure". [The other half
of the Lebesgue density theorem, namely that the limit is 0 for
almost all x NOT in E, can fail if E isn't measurable.]
Therefore, if you can show that the above limit isn't 1 at
EVERY point of your set, then your set must have measure zero.
Incidentally, this might seem to be working too hard. However,
if the limit is known to be 1 at even a single point (indeed,
if the ratios are known to have a positive lim-sup at even a
single point), then the limit will be 1 for a positive measure
set of x's.
To show that the limit isn't 1, you simply need to show that
there exists an epsilon > 0 and a sequence of R's approaching 0
(i.e. for arbitrarily small R's) such that
[measure of E intersect B(x,R)] < (1 - epsilon)*[measure of B(x,R)]
for each R in the sequence. In other words, show that the
lim-sup as R --> 0 of the ratio I gave earlier is less than 1.
In practice, it is often the case that you'll be able to show
something stronger. Namely, that for each x in your set there
exists a sequence of balls B(x, R_k) --> x such that within each
ball B(x, R_k) there exists a ball B(y_k, r_k) such that the
ratios r_k / R_k are bounded above zero. In other words, when
trying to show that the upper (i.e. lim-sup) Lebesgue density
at x is less than 1, it is often the case that there will be
sufficiently large "chunks", arbitrarily close to x and in the
complement of your set, to prevent the Lebesgue density at x
from being 1. [You don't have to scroung around looking for
sufficiently many points in B(x, R_k) missing E -- you'll
often find all you need in one of the connected components of
the complement of E.]
So here's a possible strategy ------
Since your set E is nowhere dense, we know that for each x in E
and R > 0 there exists B(y,r) contained in B(x,R) such that B(y,r)
has no points in common with E. Among all such balls B(y,r), let
B(y',r') be one of maximal size. If, for each x in E, you can find
a sequence R_k --> 0 such that r'_k/R_k is bounded above 0, then E
has measure zero. [Indeed, E will have a stronger property called
"porous" (of the lim-sup variety).]
-----------------------------------------------------------
-----------------------------------------------------------
Dave L. Renfro
Zdislav V. Kovarik
Hi,
such a set may have non-zero measure exactly because it was constructed to
do so. Follow the steps of construction, and add up the lengths of the
bounded open components of the complement.
Classical example of a thick Cantor-like discontinuum:
Take the Cantor function (if you don't know it, look it up) C(x), and
define f(x)=x+C(x).
It is continuous, strictly increasing, so it maps the Cantor set onto
another nowhere dense set on the real line, and the open intervals in the
complement into intervals of the same length.
The image of the complement of the Cantor set within [0,1] has the same
measure as the complement of the Cantor set itself, so ... (you do the
math)
Cheers, ZVK(Slavek)
William Elliot
>
> Peter <pwol...@yahoo.com> wrote:
>
>> I know that nowhere dense sets can have positive measure (such as
>> Cantor sets). If I have a specific subset of a line segment which I
>> know is nowhere dense, how can I determine it it has positive measure?
>
> At this level of generality there's really nothing that can be said.
> It has positive measure if the measure is positive.
>
> If you have a more specific set that you're wondering about someone
> may have a clue.
>
set, the middle aj-th taken from the intervals, then the
measure of the resulting fat Cantor set is prod_j (1 - aj).
Peter
> Hi,
> such a set may have non-zero measure exactly because it was constructed to
> do so. Follow the steps of construction, and add up the lengths of the
> bounded open components of the complement.
>
> Classical example of a thick Cantor-like discontinuum:
> Take the Cantor function (if you don't know it, look it up) C(x), and
> define f(x)=x+C(x).
>
it. I think in practice however, it may be difficult to implement for
3 reasons:
First, at each iteration, the function maps an infinite number of sub-
intervals (contained in the domain, which is a compact subset of R1)
to basin of attraction.
Second, the length of the sub-intervals mapping to the basin of
attraction are not the same length, nor is there a simple way to
estimate their lengths well.
Third, the function maps the domain back into itself, so there are an
infinite number of sub-intervals, after the first iteration, which do
not map into the basin of atraction, but can map into the basin of
attraction on some later iteration. Trying to account for this
behavior (even using some type of alphabet) does not seem to be
trivial.
I haven't had time to carefully go through Dave Renfro's suggestion
yet, but I'll see if I can apply that.
Thank you for the responses though,
Peter
Peter
Hi Dave,
Yes, I think this method will work for me. The mechanics underlying
this approach is to use an argument which I think is essentially what
you've just put forth.
Let E denote the set of points whose trajectories never converge.
If E is empty set, proof over, so suppose E is non-empty, therefore
there exists x* element of E.
Let f be the generating function for the dynamical system and f^n
denote the f iterated with itself n times.
I can show that the derivative of |f^n(x*)| > 1 for every natural
number, n.
This means that every open interval containing x*, no matter how
small, will upon some iterate of f map into an interval of length of
pi.
However, I know that every interval of length pi contain some sub-
interval which maps into the basin of attraction.
Hence Lim as r -> 0 of measure [E Intersect B(x*,r)] <
measure[B(x*,r)]
Thank you very much!
Peter