I notice my mistake. The problem is:
Let 0 < t < 1 be arbitrary, then there exists a measurable set E in
the set of real numbers such that den_E (0) = t. True of False? If
true, construct it. If false, prove it. (Note: den_E (0) means the
density of 0 in E.)
I revised my prove as:
True. For any 0 < t < 1. Let E = Intersection of [-tk, tk] where k is
in (0,1). Then
den_E (0)
= limit m(E Intersect [-h,h])/(2h) as h -> 0
= limit m(Intersect of [-tk,tk] where k is in (0,h])/(2h) as h -> 0
= limit (2tk)/(2k) as k -> 0
= t
I change the letter h to the letter k in the 2nd to the last line, I
don't know whether that's allowed or correct.
Hai He
--- In Math4u@yahoogroups.com, Julien Santini <julien_santini@...>
wrote:
>
> Hi Hai He,
>
> How have you been lately ? Still attending the Math
> graduate program in NY ? As for me, I came back from
> China and entered the top 1 European Business School
> for a Master of Management in Finance, but I try to
> keep on with the math sometimes !
>
> Anyway back to the problem:
>
> I thought it would work but you have too much
> variation in measure when h varies continuously.
> Here's an example: consider h=2/3^(n+1), where n is as
> large as you wish. You will then have: m(E Inter
> [-h,h])=1/3^(n+2)*1/(2/3)=1/(2*3^(n+1)), and then m(E
> Inter [-h,h])/(2h)=1/8, if I didn't commit any
> computing mistake (but then, if you pick h=1/3^n, you
> will find 1/4; therefore, the limit as h->0 does not
> exist). This tells you that the holes in between the
> components of E are too large. Hint: you need to find
> an example where your density is distributed
> "uniformly" (this would be the easiest thing to
> exhibit).
>
> --- hai_he <hai_he@...> wrote:
>
> > Define Lebesgue density of a set E (a subset of the
> > real numbers R)
> > at x as den(x) = limit m(E intersect [x-h,
> > x+h])/(2h) as h ->0 where m
> > (A) is the Lebesgue measure of a set A and [a, b] is
> > the usual
> > notation for closed interval from a to b. For
> > example, if E = [0, 1]
> > (or (0,1) or [0,1) or (0,1]), then den(0) = 1/2 =
> > den(1), den(x) = 1
> > if 0<x<1 and den(x) = 0 if x < 0 or x > 1.
> >
> > The problem is given any t with 0<t<1, does it exist
> > a set E such
> > that den(0) = t. I think it does. But I need
> > somebody to verify it:
> >
> > My goal is just to construct a set E such that
> > den(0) = t for some t
> > between 0 and 1 but not = to 1/2. If I succeed at
> > that, I can modify
> > E so that it works for any t between 0 and 1.
> >
> > My set is E = Union [2/3^(n+1), 1/3^n] as n = 0 to
> > infinity. For this
> > E, clearly m(E) = 1/2. Therefore, as h -> 0,
> > lim m(E intersect [-h, h])/(2h)
> > = lim m(E intersect [0, h])/(2h)
> > = lim ((1/2)*h)/(2h)
> > = 1/4
> >
> > My problem is I can't convince myself that m(E
> > intersect [0, h]) is
> > (1/2)*h, however I think it is when h -> 0. I just
> > need somebody to
> > show me my above limit is correct with a stronger
> > reason.
> >
> > Thank you.
> >
> > Hai He
> >
> >
> >
>
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