nowhere dense + measure zero
Robert Valkenburg
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?
Thanks.
regards
Robert Valkenburg
klaus hoffmann
for arbitrary small epsilon (see J.C. Oxtoby, "Measure and Category" Springer
Graduate Text in Mathematics 2, pp. 4 f, isbn 0-387-05349-2).
So I would assume that nowhere dense is not very helpful for showing "measure zero"
Klaus
Robert Valkenburg schrieb:
Daniel Luecking
>
> Hi,
>
> 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?
> Thanks.
There really aren't any conditions that are better than the definition:
they are be contained in a union of rectangle with arbitrarily small
total measure.
Certainly nowhere dense is necessay for measure zero, but nowhere dense
subsets of a rectangle can have measure arbitrarily close to the measure
of the rectangle (even equal if the set isn't closed).
>From another direction, there is unlikely to be any reasonable
topological criterion, since any nowhere dense closed set can be shown
to be homeomorphic to a set of measure zero.
Dave L. Renfro
[sci.math.research Mon, 19 Feb 2001 17:13:46 +1300]
<http://forum.swarthmore.edu/epigone/sci.math.research/suntwatwi>
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