An unusual measurable set
TCL
f(r)= m(E \cap I)/m(I) satisfies
liminf_{r>0} f(r)=0 and limsup_{r>0} f(r)=1. Here I=(-r,r) and m is the
Lebesgue
measure.
A N Niel
Construct it as a countable union of intervals, clustering at 0.
Jules
How about this: Let E = {x : 1 / n! > |x| > 1 / (n + 1)! for some even
n}. Then, for even n, f(1 / n!) > n / (n + 1), and for n odd, f(1 /
n!) < 1 / (n + 1).
The World Wide Wade
"TCL" <tl...@verizon.net> wrote:
> I am trying to find a (Lebesgue) measurable set E such that the function
> f(r)= m(E \cap I)/m(I)
no r on the RHS?
> satisfies
> liminf_{r>0} f(r)=0 and limsup_{r>0} f(r)=1. Here I=(-r,r) and m is the
> Lebesgue
> measure.
Show there is a sequence b_1 > a_1 > b_2 > a_2 > ... -> 0, such
that (b_n - a_n)/b_n -> 1 and a_n/b_(n+1) -> 0. Set E = U {a_n <
|x| < b_n}, and consider f(b_n) and f(a_n).
TCL
"The World Wide Wade" <wadera...@comcast.remove13.net> wrote in message
news:waderameyxiii-188...@comcast.dca.giganews.com...
> In article <3UF7h.827$a_2.159@trnddc01>,
> "TCL" <tl...@verizon.net> wrote:
>
>> I am trying to find a (Lebesgue) measurable set E such that the function
>> f(r)= m(E \cap I)/m(I)
>
> no r on the RHS?
>
>> satisfies
>> liminf_{r>0} f(r)=0 and limsup_{r>0} f(r)=1. Here I=(-r,r) and m is the
>> Lebesgue
>> measure.
>
> Show there is a sequence b_1 > a_1 > b_2 > a_2 > ... -> 0, such
> that (b_n - a_n)/b_n -> 1 and a_n/b_(n+1) -> 0.
I think you meant S_n / b_n ->1 and S_n / a_{n-1}-->0, where
S_n = sum_{k=n}^\infty (b_k - a_k).
Jules' example does that.
What if we replace 1 by d, 0 by c, where 0<=c<=d<=1, do you think one
can still find such sequences a_n, b_n?
The World Wide Wade
"TCL" <tl...@verizon.net> wrote:
> "The World Wide Wade" <wadera...@comcast.remove13.net> wrote in message
> news:waderameyxiii-188...@comcast.dca.giganews.com...
> > In article <3UF7h.827$a_2.159@trnddc01>,
> > "TCL" <tl...@verizon.net> wrote:
> >
> >> I am trying to find a (Lebesgue) measurable set E such that the function
> >> f(r)= m(E \cap I)/m(I)
> >
> > no r on the RHS?
> >
> >> satisfies
> >> liminf_{r>0} f(r)=0 and limsup_{r>0} f(r)=1. Here I=(-r,r) and m is the
> >> Lebesgue
> >> measure.
> >
> > Show there is a sequence b_1 > a_1 > b_2 > a_2 > ... -> 0, such
> > that (b_n - a_n)/b_n -> 1 and a_n/b_(n+1) -> 0.
>
> I think you meant S_n / b_n ->1 and S_n / a_{n-1}-->0, where
> S_n = sum_{k=n}^\infty (b_k - a_k).
No, I meant what I wrote, except for a typo: a_n/b_(n+1) should
have been b_(n+1)/a_n. This easily implies anything you need to
know about S_n here.
Dave L. Renfro
Others have commented on your specific question, so I thought
I'd mention some extensions. You're looking at the upper (limsup)
and lower (liminf) symmetric Lebesgue densities of E at the
point 0. There are also the ordinary upper and lower Lebesgue
densities of a set at a point, which are defined analogously
by using a limsup and a liminf over all open intervals I
containing the point. For a given measurable set and a given
point, let LD^ and LD_ be the upper and lower ordinary Lebesgue
densities, and let SLD^ and SLD_ be the upper and lower symmetric
Lebesgue densities for the set at that point. Clearly, for each
such set and point, we have
0 <= LD_ <= SLD_ <= SLD^ <= LD^ <= 1.
The Lebesgue density theorem says that for each measurable
set E, almost every point (i.e. all but a Lebesgue measure
zero set of points) on the real line belongs to
P union Q,
where
P = {x: LD_ = SLD_ = SLD^ = LD^ = 0}
Q = {x: LD_ = SLD_ = SLD^ = LD^ = 1}
In particular, LD_ = SLD_ = SLD^ = LD^ almost everywhere.
In fact, LD^ = 0 a.e. outside of E and LD_ = 1 a.e. in E.
(The latter, but not always the former, holds even if
E is not measurable, if we use the outer measure analogs.)
Theorem 1 in [Goffman] (see below) states that given
any measure zero set Z, there exists a measurable
set E (in fact, E can be chosen to be an F_sigma set)
such that for the set E, LD_ differs from LD^ at each
point of Z. The *proof* of Goffman's Theorem 1 actually
shows that for the set E we have, at each point of Z,
LD_ = 0 and LD^ = 1.
However, we cannot conclude from this that for the set E
we have, at each point of Z, SLD_ = 0 and SLD^ = 1 (see the
inequality chain displayed earlier), or even that there
exists a measurable set such that SLD_ differs from SLD^
at each point of Z. Nonetheless, with a bit more care
in Goffman's proof, we can show there exists an F_sigma
set E such that for the set E we have, at each point of Z,
SLD_ = 0 and SLD^ = 1.
Casper Goffman, "On Lebesgue's density theorem", Proceedings
of the American Mathematical Society 1 (1950), 384-388.
Dave L. Renfro