Letting $delta rightarrow infty$ in the $delta$-Hausdorff premeasure

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This may come off as a naive question but it's something I've always wondered.
We define the Hausdorff measure as $mathcal H^s(A) = lim_deltarightarrow0mathcal H^s_delta(A)$ for some $Asubseteqmathbb R^n$ where $mathcal H^s_delta(A) = infleftlbrace sum_alpha left(fracmathrmdiam(E_alpha)2right)^s rightrbrace$ where $(E_alpha)$ is a finite or countable cover of $A$ such that $mathrmdiam(E_alpha) < delta$.



According to intuition, decreasing the diameter of the covering sets should make the measurement more precise so I understand why the measure is defined as it is. However it's easy to prove that $mathcal H^s_delta$ is decreasing in $delta$ so I've been wondering if there's any sense in defining $h^s(A) = lim_deltarightarrowinftymathcal H^s_delta(A)$.



If I'm not mistaken $h^s(A)$ should always be finite, is there any way I can interpret its value?




measure-theory hausdorff-measure




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    This may come off as a naive question but it's something I've always wondered.
    We define the Hausdorff measure as $mathcal H^s(A) = lim_deltarightarrow0mathcal H^s_delta(A)$ for some $Asubseteqmathbb R^n$ where $mathcal H^s_delta(A) = infleftlbrace sum_alpha left(fracmathrmdiam(E_alpha)2right)^s rightrbrace$ where $(E_alpha)$ is a finite or countable cover of $A$ such that $mathrmdiam(E_alpha) < delta$.



    According to intuition, decreasing the diameter of the covering sets should make the measurement more precise so I understand why the measure is defined as it is. However it's easy to prove that $mathcal H^s_delta$ is decreasing in $delta$ so I've been wondering if there's any sense in defining $h^s(A) = lim_deltarightarrowinftymathcal H^s_delta(A)$.



    If I'm not mistaken $h^s(A)$ should always be finite, is there any way I can interpret its value?




    measure-theory hausdorff-measure




    share|cite|improve this question





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      This may come off as a naive question but it's something I've always wondered.
      We define the Hausdorff measure as $mathcal H^s(A) = lim_deltarightarrow0mathcal H^s_delta(A)$ for some $Asubseteqmathbb R^n$ where $mathcal H^s_delta(A) = infleftlbrace sum_alpha left(fracmathrmdiam(E_alpha)2right)^s rightrbrace$ where $(E_alpha)$ is a finite or countable cover of $A$ such that $mathrmdiam(E_alpha) < delta$.



      According to intuition, decreasing the diameter of the covering sets should make the measurement more precise so I understand why the measure is defined as it is. However it's easy to prove that $mathcal H^s_delta$ is decreasing in $delta$ so I've been wondering if there's any sense in defining $h^s(A) = lim_deltarightarrowinftymathcal H^s_delta(A)$.



      If I'm not mistaken $h^s(A)$ should always be finite, is there any way I can interpret its value?




      measure-theory hausdorff-measure




      share|cite|improve this question











      This may come off as a naive question but it's something I've always wondered.
      We define the Hausdorff measure as $mathcal H^s(A) = lim_deltarightarrow0mathcal H^s_delta(A)$ for some $Asubseteqmathbb R^n$ where $mathcal H^s_delta(A) = infleftlbrace sum_alpha left(fracmathrmdiam(E_alpha)2right)^s rightrbrace$ where $(E_alpha)$ is a finite or countable cover of $A$ such that $mathrmdiam(E_alpha) < delta$.



      According to intuition, decreasing the diameter of the covering sets should make the measurement more precise so I understand why the measure is defined as it is. However it's easy to prove that $mathcal H^s_delta$ is decreasing in $delta$ so I've been wondering if there's any sense in defining $h^s(A) = lim_deltarightarrowinftymathcal H^s_delta(A)$.



      If I'm not mistaken $h^s(A)$ should always be finite, is there any way I can interpret its value?





      measure-theory hausdorff-measure





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