$\delta$-approximative-p-dimensional Hausdorff measure a measure?

Question

Might someone explain why it is necessary to define the Hausdorff-measure as the limit of the outer measure depictured below?

Answer 1

One possible answer to that is dimensionality: If you pick an arbitrary compact set $K$, then $\mathcal{H}^p_\delta(K)<\infty$. This can be seen by picking a cover of $K$ of open balls with diameter less than $\delta$. Then by compactness you find a finite numer of these balls still covering $K$. Hence they yield an upper bound on $\mathcal{H}^p_\delta(K)$.

Now pick a closed square $Q$ in $\mathbb{R}^2$, for example $Q:=[0,1]\times[0,1]$. This should be a set of dimension two and therefore a $1$-dimensional area measure should yield $\infty$. But $\mathcal{H}^1_\delta(Q) <\infty$ as seen above.

Hence $\mathcal{H}^p_\delta$ does not give you the right term of dimensionality, but $\lim_{\delta\searrow 0}\mathcal{H}^p_\delta$ yields exactly that.

EDIT: $\mathcal{H}^p_\delta$ is in general not a Borel measure. Let me give you an example: For a straight line $L\subset \mathbb{R}^2$ you can show $\mathcal{H}^1_\delta(L)=\mathcal{H}^1(L) = \mathcal{L}^1(A\circ L)$. Here $A$ is a rotation and translation, such that $L\subset \mathbb{R}\times\{0\}=\mathbb{R}$, i.e. an isometry. Now let $A=[0,1]\times\{0\}$ and $B=[0,1]\times\{\varepsilon\}$ with $\varepsilon>0$. Then $$\mathcal{H}^1_\delta (A) = \mathcal{H}^1_\delta(B)=1$$ hence $$\mathcal{H}^1_\delta(A) + \mathcal{H}^1_\delta(B)=2,$$ but if you choose $\delta>0$ big enough, $A\cup B$ cover $A\cup B$ and $\operatorname{diam}(A\cup B) < \delta$, but $\operatorname{diam}(A\cup B)=\sqrt{1+\varepsilon^2}<2$ if $\varepsilon$ is small enough. Hence $$H^1_\delta(A\cup B) <2\mbox{ and } \mathcal{H}^1_\delta(A) + \mathcal{H}^1_\delta(B)=2$$ Since $A$ and $B$ are Borel sets, $\mathcal{H}^1_\delta$ is not a Borel measure.