let $D=\{(\frac{i}{p},\frac{j}{p}:p\in \mathbb N, i,j=1,2,...,p-1\}$. Does $D$ has content zero? Definition of content zero: $D\subset \mathbb R^2$ if for every $\epsilon>0$ there exists a finit collection of rectangle $R_k$, $1\leq k\leq n$ whose union cover $D$ and the sum of their area is less than $\epsilon$.
Let $x=\frac{i}{p}$ and $y=f(x)=\frac{j}{p}$. From the definition of $D$, we can see that rational numbers $x,y \in (0,1]$ I know that $f(x)$ is continuous on its domain, but its domain is not compact, so I can't argue that $f(x)$ is uniformly continuous, and use the definition of uniformly continuous to move forward and establish the rectangle. I even have the feeling that this set doesn't have content zero, but I don't know how to prove that.
 A: Note that $D$ is dense in $[0,1]\times [0,1]$. Suppose the union of $R_k$ covers $D$. We can assume WLOG that the $R_k$ are closed, so since there are finitely many their union is closed. Thus their union contains the closure of $D$, which is $[0,1]\times [0,1]$, so the sum of the areas of the $R_k$ must be at least $1$. 
Thus $D$ does not have content zero.
Edit: To see that the sum of the areas must be at least $1$, subdivide $[0,1]\times [0,1]$ into rectangles of the form $[a2^{-n},(a+1)2^{-n}]\times [b2^{-n},(b+1)2^{-n}]$. Take $n$ large enough that $2^{-n}$ is less than half the smallest overlap between any pair of intersecting rectangles among the $R_k$, thus every $[a2^{-n},(a+1)2^{-n}]\times [b2^{-n},(b+1)2^{-n}]$ is properly contained in some $R_k$. The area of each $R_k=[a_k,b_k]\times [c_k,d_k]$ is at least the sum of the areas of these rectangles contained in it, since these rectangles just partition $[a_k',b_k']\times [c_k',d_k']$ for some $a_k'> a_k,b_k'<b_k,c_k'>c_k$ and $d_k'<d_k$. Thus the sum of the areas of the $R_k$ is at least the sum of the areas of the $[a2^{-n},(a+1)2^{-n}]\times [b2^{-n},(b+1)2^{-n}]$, which is $1$.
