Calculus Apostol Exercise 1.11 number 6 
I just don't know how to start with this problem, it seems obvious but I'm just stumped on how to start the proof. Any suggestions?
 A: We can help ourselves by drawing a picture to get a good idea of what is going on, then turn that intuition into something more rigorous.  The picture is as follows:

$\hskip1.5in$ 

In the picture, we can see (the filled in dots) the lattice points in the ordinate set of $f(x)$ (not including the $x$-axis since the question stipulates $S$ contains the points $0 < y \leq f(x)$).  At each integer between $a$ and $b$, we count the number of lattice points beneath $[f(x)]$, the greatest integer less than or equal to $f(x)$.  Then we need to turn this intuition from the picture into a proof:
 Proof.  Let $n \in \mathbb{Z}$ with $a \leq n < b$.  We know such an $n$ exists since $a,b \in \mathbb{Z}$ with $a < b$.  Then, the number of lattice points in $S$ with first element $n$ is the number of integers $y$ such that $0 < y \leq f(n)$.  But, by definition, this is $[f(n)]$ (the greatest integer less than or equal to $f(n)$).  Summing over all integers $n, \ a \leq n \leq b$ we have,
$$ \text{# of lattice points in } S = \sum_{n=a}^b [ f(n) ]. \qquad \blacksquare$$
