Re: Lebesgue differentiation theorem
Yihong
will the general theory of differentiation of measures (on R) stay the same if we replace in our definitions (e.g. density) every [x-r, x+r] by [x-a*r,x+(1-a)*r] for some fixed 0 < a < 1? will the Radon-Nikodym theorem still holds?
> Dear all,
>
> I have a question about Lebesgue-Besicovitch
> differentiation theorem as follows:
>
> The original theorem states that: if f belongs to
> L^1_{loc}(R, \mu), where \mu is a Radon measure, then
> for \mu-a.e. x,
>
> lim_{r -> 0} \mu[x-r,x+r]^{-1} \int_{[x-r,x+r]} f d
> \mu = f(x).
>
> My question is that *can we replace [x-r,x+r] by
> [x-a*r,x+(1-a)*r] for some fixed 0 < a < 1, i.e., the
> balls are not centered by are of a fixed
> eccentricity?*
>
> My remarks are as follows:
> 1. If \mu is the Lebesgue measure, then the answer is
> true. In fact, the limit can be taken on any
> shrinking balls containing x, as shown in many
> books.
>
> 2. I followed the approach in Federer's Geometric
> Measure Theory [2.8.17] to check that {(x,
> [x-a*r,x+(1-a)*r])} is a Vitali relation, and hence
> the conclusion holds by [2.9.9]. I wonder if there is
> a direct approach w/o resorting to the general theory
> of Federer.
>
> Thanks,
> Yihong
Dave L. Renfro
> will the general theory of differentiation of measures
> (on R) stay the same if we replace in our definitions
> (e.g. density) every [x-r, x+r] by [x-a*r,x+(1-a)*r]
> for some fixed 0 < a < 1? will the Radon-Nikodym theorem
> still holds?
I know very little about the abstract theory (general theory
of differentiation of measures), but I do know that if you're
considering a situation in which only "density = 0" and
"density = 1" are relevant, then there is no change when
an arbitrary (but fixed for the process) choice of an
unsymmetrical interval type is used. In fact, this continues
to be true when the choice of an unsymmetrical interval type
is allowed to vary when approaching the point and, in addition,
when the way the unsymmetrical interval type varies as you
approach a point does not have to be uniform with respect
to the point (this most general situation is when upper and
lower density are defined by using a limsup and a liminf
over all open intervals containing the point).
I don't know how relevant what follows is to your question,
but here are some comments about Lebesgue density I posted
a few years ago.
************************************************************
sci.math -- An unusual measurable set (21 November 2006)
http://groups.google.com/group/sci.math/msg/5713e0d9547e6cdc
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.
************************************************************
Incidentally, to add to what I said in this 2006 post, the
following is known (see [1]):
For each subset E of the reals, the set of points in R
such that
SLD+ = SLD- and LD+ 'not equal' LD-
is sigma-porous in R.
[1] This is given in Theorem 5 (p. 266) of Charles L. Belna,
Michael J. Evans, and Paul D. Humke, "Symmetric and ordinary
differentiation", Proceedings of the American Mathematical
Society 72 (1978), 261-267.
Dave L. Renfro