0
$\begingroup$

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?.

$\endgroup$
  • $\begingroup$ What is $I{}{}$? $\endgroup$ – Lord Shark the Unknown Oct 7 '18 at 17:17
  • $\begingroup$ sorry. $I$ is an interval. $\endgroup$ – JohanR Oct 7 '18 at 17:20
  • $\begingroup$ you need $\frac{1}{N}$ in front of the count in the definition of the limit. $\endgroup$ – Hayk Oct 13 '18 at 16:26
0
$\begingroup$

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 $N \to \infty$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.