I'm reading the book about measure theory of Terry Tao by my own, and I want to prove that: $$ | I| = \lim_{N \to \infty} \# \left( I \cap \frac{Z}{N} \right)$$
where $I$ is an interval, $\#A$ denotes the cardinality of $A$, and $\frac{Z}{N}=\left\{\frac{n}{N}: n \in Z \right\} $.
If I define a sequence like: $a_{N}=\frac{1}{N}\#\left( I \cap \frac{Z}{N} \right)$, what is its behavior?.
Let $I = [a,b]$ with $a<b$ reals (it does not matter if the endpoints are included or not as the argument below shows). Then for $N\in \mathbb{N}$ we have $$ \# \left( I \cap \frac{Z}{N} \right) = \#\{n \in \mathbb{Z}: a N \leq n \leq b N\} := a_N. $$
Then $$ [bN] - [aN] \leq a_N \leq [bN] - [aN] + 1 \tag{1} $$ where $[\cdot]$ denotes the integer part of a real number. For each $x\in \mathbb{R}$, in view of the definition of the integer part we have $xN - 1 < [xN] \leq xN$ and hence $$ x - \frac{1}{N} < \frac{[xN]}{N} \leq x , $$ which implies that $\lim\limits_{N\to \infty} \frac{[xN]}{N} = x$. Applying this on $(1)$ shows that $\frac{a_N}{N} \to b-a$ as $N \to \infty$.