Hausdorff content and Hausdorff measure I am dealing with the Hausdorff dimension and I came across two different ways of defining this dimension. This question is possibly related to Hausdorff Measure and Hausdorff Dimension but the answers there don't completely take my doubts away.
One definition uses the Hausdorff content, as stated on wikipedia http://en.wikipedia.org/wiki/Hausdorff_dimension, and the other one --which as far as I am concerned is the canon one-- uses the Hausdorff measure http://en.wikipedia.org/wiki/Hausdorff_measure.
Now, this puzzles me because no matter how I look at it, both the Hausdorff content and the Hausdorff measure look the same to me. What am I missing?
If someone could provide a reference about this matter I'd be so happy. I am particularly interested in a text which deals on the relationship between the Hausdorff dimension and the Box-counting dimension, especially the cases where they match.
 A: First, Wikipedia does not always follow definitions used elsewhere...  But for the purposes in the cited pages:
In "Hausdorff content" we do not have a $\delta \to 0$ clause in there, but for "Hausdorff measure" we do have the $\delta \to 0$ clause.  In order to define "Hausdorff dimension", the claim is that we can do it either way, and get the same result.  The two have the same "measure zero" sets.  But, in fact, the "Hausdorff measure" and "Hausdorff content" of a set can be different (as long as they are not zero).
Example. 
In $\mathbb R$, consider the set $E=[0,1]$ and investigate dimension $d=1/2$.  There is a cover by a set with diameter $1$, namely the single set $[0,1] = B(1/2,1/2)$, so the $1/2$-dimensional Hausdorff content is at most $1$, $C^{1/2}_H(E) \le 1$.
But if you cover $E$ by sets of diameter $\le \delta=1/n$, then (it turns out) you cannot do better than the cover by $n$ balls of diameter $1/n$, so that
$$
H^{1/2}_{1/n}(E) \ge n^{1/2},
$$
and the Hausdorff measure itself is $\infty$, the limit of $H^{1/2}_{\delta}(E)$ as $\delta \to 0^+$.
A: Here is some direct proof. Assume H^k-content of a set is zero. Then by the definition there exists a covering by sets such that the sum of diameters^k is small. Now, although definition of H_content does not require any bound on the diameters, the fact that this latter sum is small implies that all diameters are (voluntarily) small. Thus, we can use this same covering in the definition of H_delta, for small delta. And letting delta go to zero (each time you find a new covering), we see that H^k is also zero for the set. So, we proved: H_content = 0 $\implies$ H^k=0.
