Elias Stein : Real Analysis I cannot understand why this particular line in the text is true: 
" Moreover, there are $O(k^{d-1})$ cubes in  $\cal{Q}\ '$ "
For the text see 
http://books.google.com/books?id=2Sg3Vug65AsC&pg=PA12&lpg=PA12&dq=elias+stein+Moreover,+there+are+O%28k+d%C2%A11+%29+cubes+2+in+Q+0&source=bl&ots=DCMQaf06nz&sig=1V3GKoVbY9Xq8uOdJtD7kpYS9ug&hl=en&sa=X&ei=UQYgVO7yH5GayAS_1IL4Bw&ved=0CCAQ6AEwAA#v=onepage&q=elias%20stein%20Moreover%2C%20there%20are%20O%28k%20d%C2%A11%20%29%20cubes%202%20in%20Q%200&f=false

In the text we have the following situation. Consider a rectangle $R$ in $\mathbb{R}^d$. Fix a grid of cubes of side length $1/k$ for $k\in\mathbb{N}$ large.
Let $\mathcal{Q}^\prime$ be the collection of grid cubes $Q$ such that $Q\cap R\not=\emptyset$ and $Q\cap R^c\not=\emptyset$ (it is formulated a bit ambiguously in the text, but this is clearly what is meant). That is, $\mathcal{Q}^\prime$ consists of the grid cubes that intersect the boundary of $R$. Note that the boundary of $R$ is a union of $(d-1)$-dimensional rectangles.
One grid cube intersects the boundary on a surface of size comparable to $(1/k)^{d-1}$. Thus to cover the entire boundary we need about $\frac{1}{(1/k)^{d-1}}=k^{d-1}$ grid cubes, up to a constant depending on the size of $R$. That is, $\mathcal{Q}^\prime$ contains $O(k^{d-1})$ cubes.
