Sufficient condition for integer Hausdorff dimension. It is pretty much in the title: is there a non-trivial sufficient condition on geometrical shapes that forces the Hausdorff dimension to be an integer ?
Most fractals look "complicated" in some way, so it seems they should violate some strong regularity condition.
I am not familiar with the topic so I apologize if it is an obvious question, but I could not find an answer by googling. Any reference will be appreciated.
 A: Yes.  A notion of regularity can be stated in terms of density in a way that is analogous to the Lebesgue density theorem.  For this to work, it turns out that the dimension must be an integer.  The following details are based on section 5.1 of Falconer's Fractal Geometry.
Given a set $F\subset\mathbb R^n$ of positive and finite $s$-dimensional Hausdorff measure, let us define the upper and lower densities of $F$ at a point $x\in\mathbb R^n$ by
$$
\overline{D}^s(F,x) = \limsup_{r\rightarrow 0} \frac{{\cal H}^s(F\cap B_r(x))}{(2r)^s}
$$
and
$$
\underline{D}^s(F,x) = \liminf_{r\rightarrow 0} \frac{{\cal H}^s(F\cap B_r(x))}{(2r)^s}.
$$
A point at which the upper and lower densities are both 1 is called regular and a point which is not regular is called irregular.  $F$ is called regular if $\cal H^s$ almost all of its points are regular.  Theorem 5.2 of the cited text states:
$F$ is irregular, unless $s$ is an integer.
In particular, a regular set must have integer dimension.  Note that the converse is not true.  There are irregular sets in the plane of positive, finite, one-dimensional Hausdorff measure.  A Sierpinski-like set with the pieces scaled by the factor $1/3$ would be an example.
