Quote:
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but showing the Hausdorff Measure is infinity for dimension less than 1...
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Consider just the half interval
.
Let's suppose that
is finite for some
. Choose a covering
of
costisting of intervals of individual length
. Their number must be no less than
![\left[1/\delta\right]](http://www.mathhelpforum.com/math-help/latex2/img/9168cc7cfa463afa2fb4a7f49d225d69-1.gif "\left[1/\delta\right]")
(as
is also a covering for the Lebesgue measure of I in
). This means that the related Hausdorff sum will be equal to
![C\delta^{1-\lambda}\left[1/\delta\right]](http://www.mathhelpforum.com/math-help/latex2/img/de84c5a7ccee516873f88d211320f53e-1.gif "C\delta^{1-\lambda}\left[1/\delta\right]")
, where
is a constant not depending on
or
. So, the infimum
of all such partitions (which we assumed exists) satisfies
![m(\delta)\geq C\delta^{1-\lambda}\left[1/\delta\right]](http://www.mathhelpforum.com/math-help/latex2/img/a4063fede3eff7634bbfbc0173abfa44-1.gif "m(\delta)\geq C\delta^{1-\lambda}\left[1/\delta\right]")
. Now in the last relation, the limit of the right hand side as
behaves like the limit of
as
, which is infinite. A contradiction.