Prove carpet has positive Hausdorff measure in its dimension 
Given $D\subset\{0,1,2,\dots n-1\}\times\{0,\dots,m-1\}$, let $$K(D)=\{\sum_{k=1}^\infty(a_kn^{-k},b_km^{-k}):(a_k,b_k)\in D\forall k\}.$$
  Show that if $D$ has uniform horizontal fibers (i.e. the number of rectangles in each row of $K(D)$ is equal), it means that $K(D)$ has positive Hausdorff measure in its dimension.

I find the formulation of this question a bit unclear I know that suppose $K(D)$ has uniform fibers the Malinowski and Hausdorff dimensions agree and $$\dim K(D)=1+\log_n\frac{\#D}{m}$$(where #D is the cardinality of D) but I cannot understand how it implies that exist a positive Hausdorff measure?
 A: The following lemma is theorem 5.8 of Falconer's The Geometry of Fractal Sets and, I believe, is directly applicable to your situation.

Let $F$ be a subset of the $xz$ plane, let $A$ be any subset of 
the $z$-axis, and for $z\in A$, let
$$F_z = \{x: (x,z)\in F\}.$$
Suppose there is a constant $c>0$ such that ${\cal H}^{t}(F_z) > c$ for all $z\in A$. Then 
$${\cal H}^{s+t}(F) \geq b \; c \; {\cal H}^s(A),$$
where $b$ depends only on $s$ and $t$.

Examining your specific question in a bit more detail, the notation is, indeed, a bit difficult.  It was introduced in a paper by Curt McMullen, though, so who am I to complain?  Let's examine a couple of simple examples to help us figure it out.  
First, let's look a simpler, one-dimensional example.  Suppose that $n=3$ and $D=\{0,2\}$  The one-dimensional analogue of your sum set is then 
$$\left\{x=\sum_{k=1}^{\infty} \frac{a_k}{3^k}: a_k = 0 \text{ or } 2\right\}.$$
This is a well known characterization of the Cantor set.  
The Cantor set, of course, is often described as a self-similar set, in that it is made of two copies of itself scaled by the factor $1/3$.  Your set $K(D)$ is self-similar only when $m=n$ but, more generally, is self-affine; it is made of smaller copies of itself scaled by the factor $1/n$ horizontally and $1/m$ vertically.  To make this clear, let is again look at an example.  Let's suppose that $n=6$, $m=3$, and that
$$D=\{\{0, 0\}, \{1, 0\}, \{3, 0\}, \{5, 0\}, \{1, 1\}, \{2, 1\}, \{4, 1\}, \{5, 1\}, \{0, 2\}, \{1, 2\}, \{4, 2\}, \{5, 2\}\}.$$
Then,
$$K(D)=\left\{(x,y) = \sum_{k=1}^\infty(a_kn^{-k},b_km^{-k}): (a_k,b_k)\in D\right\}$$
can be visualized like so:

The rectangle labeled $(i,j)$ is the set of all points in $K(D)$ such that $(a_1,b_1)=(i,j)$.  We can iterate the procedure to obtain a better visualization of $K(D)$.  After several iterates, we get something like so:

Note that we have the same number of rectangles in each row. That's the manifestation of the "uniform horizontal fibers" which allows us to apply the lemma.
