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There are subsets of the real line that has infinite counting measure, but Lebegue measure 0, so the Lebegue measure is used for measuring larger sets than the counting measure. My question is: Is there a translation invariant measure m such that for some sets with Lebegue measure 0 the m-measure is infinite and for some sets with infinite counting measure, the m-measure is 0?

I have found one example: m(A)=0 if A is countable, and m(A)=infinite otherwise. So I will require that the measure can take the value 1.

If such a measure exist, can we find a measure between this and the counting measure? and between this and the Lebegue measure? and so on.

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2 Answers 2

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Hausdorff measures of dimensions between 0 and 1 are a continuous spectrum of examples.

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  • $\begingroup$ And I guess adding Hausdorff measures with gauge functions would give even more examples, which aren't all that easy to works with though. $\endgroup$
    – Thorny
    Commented Jan 13, 2010 at 9:24
  • $\begingroup$ Acknowledgment: My response is independent of Thorny's but came slightly later. $\endgroup$
    – Pete L. Clark
    Commented Jan 13, 2010 at 9:27
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    $\begingroup$ the conditions require that there exists a set with lebesgue measure 0 but infinite measure according to this measure. Are you sure this holds for Hausdorff measure (with dimension in $(0,1)$)? $\endgroup$
    – Matus Telgarsky
    Commented Jan 13, 2010 at 9:41
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    $\begingroup$ If a set has positive, finite d-dimensional measure (and it is easy to create for example a self-similar Cantor set with a prescribed Hausdorff dimension), then it has measure 0 according to any d'-dimensional Hausdorff measure with d'>d and has infinite measure according to any d"-dimensional Hausdorff measure with d"<d. $\endgroup$
    – Thorny
    Commented Jan 13, 2010 at 10:26
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I believe what you are looking for is the $\alpha$-dimensional Hausdorff measure for some $0 < \alpha < 1$.

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