Re: [Math4u] Lebesgue density

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@yahoo.com> 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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