# what is the limit and behavior of this sequence?

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?.

• What is $I{}{}$? – Lord Shark the Unknown Oct 7 '18 at 17:17
• sorry. $I$ is an interval. – JohanR Oct 7 '18 at 17:20
• you need $\frac{1}{N}$ in front of the count in the definition of the limit. – Hayk Oct 13 '18 at 16:26

Let $$I = [a,b]$$ with $$a reals (it does not matter if the endpoints are included or not as the argument below shows). Then for $$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$$.