Recovering length of an interval by a limiting formula This is from Tao's text on Measure theory. While proving Lemma 1.1.2-(ii) he's using a discretization argument for length of an interval in $\mathbb{R}$, which is : $$|I|:=\lim_{N\to \infty}\frac{1}{N}\operatorname{card}\left(I \cap \frac{1}{N}\mathbb{Z}\right)$$ Where $\frac{1}{N}\mathbb{Z}:=\left\{\frac{n}{N}:n \in \mathbb{Z}\right\}$, and by $\operatorname{card}$ i mean cardinality of a finite set. I do have an intuitive idea for this, but not able to prove the result. I was trying like this: for any $I$ say $I:=[-5,5]$, if i take some element $i \in [-5,5]\cap\frac{1}{N}\mathbb{Z}$, then $i=j/N:j\in \mathbb{Z}\cap [-5,5]$ and then $\operatorname{card}\left([-5,5] \cap \frac{1}{N}\mathbb{Z}\right)=|(-5)N-(5)N|$, afterthat i'm not sure how to finish this argument.
 A: I think i get where i was going wrong, I'll add it as an answer.
Let, $I:=[a,b]$ and $I \cap \frac{\mathbb{Z}}{N}:=\left\{\frac{k}{N}\in I: k\in \mathbb{Z}\right\}$. This exactly implies that $a \le \frac{k}{N}\le b \implies Na \le k\le Nb$. So cardinality of the above set is the number of integers satisfying the inequality.
$$\operatorname{card}\left(I \cap \frac{\mathbb{Z}}{N}\right)=\lfloor Nb\rfloor-\lceil Na \rceil +1$$Now we have : $ \lfloor Nb\rfloor \le Nb$ and $\lceil Na \rceil \ge Na$, so $$\lfloor Nb\rfloor-\lceil Na \rceil +1 \le Nb-Na+1=N(b-a)+1$$ which implies $$\frac{\lfloor Nb\rfloor-\lceil Na \rceil +1}{N}\le (b-a)+\frac{1}{N}$$ moreover, $\lfloor Nb\rfloor \ge Nb-1$ and $\lceil Na \rceil \le Na+1$ which implies $$\frac{\lfloor Nb\rfloor-\lceil Na \rceil +1}{N}\ge \frac{(Nb-1)-(Na+1)+1}{N}=(b-a)-\frac{1}{N}$$
Finally, we write  $$(b-a)-\frac{1}{N} \le \frac{\lfloor Nb\rfloor-\lceil Na \rceil +1}{N}\le (b-a)+\frac{1}{N}$$ Now letting $N\to \infty$ gives us the desired result: $$\lim_{N\to \infty}\frac{\lfloor Nb\rfloor-\lceil Na \rceil +1}{N}=b-a=m(I)$$
A: Actually, it occurred to me that there is an error in your argument. It is not necessarily the case that $\text{card}(I\bigcap\mathbb{Z})=\lfloor Nb\rfloor-\lceil Na\rceil+1$. Fortunately, you need not correct for this error when you observe the simpler argument.
Notice that $I\bigcap\frac{1}{N}\mathbb{Z}=\{x\in I|x=\frac{k}{N}, k\in\mathbb{Z}\}$ and $|I|=b-a$. Therefore, the cardinality of $I\bigcap\frac{1}{N}\mathbb{Z}$ is the number of elements $x=\frac{k}{N}$ such that $a\leq x\leq b$. Equivalently, the cardinality of $I\bigcap\frac{1}{N}\mathbb{Z}$ is the number of integers $k$ such that $aN\leq k\leq bN$; it is the quantity $\lfloor{bN-aN}\rfloor+1=\lceil{bN-aN}\rceil+1$. Therefore,
$\lfloor{bN-aN}\rfloor+1=\lceil{bN-aN}\rceil+1=\textit{card}\left(I\bigcap\frac{1}{N}\mathbb{Z}\right)$.
So it follows that,
\begin{align*}
  \lfloor{b-a}\rfloor&=\lim\limits_{N\rightarrow\infty}{\frac{N}{N}\left(\lfloor{b-a}\rfloor+\frac{1}{N}\right)}\\
  &=\lim\limits_{N\rightarrow\infty}{\frac{1}{N}\left(\lfloor{bN-aN}\rfloor+1\right)}\\
  &=\lim\limits_{N\rightarrow\infty}{\frac{1}{N}\left(\lceil{bN-aN}\rceil+1\right)}\\
  &=\lim\limits_{N\rightarrow\infty}{\frac{N}{N}\left(\lceil{b-a}\rceil+\frac{1}{N}\right)}\\
  &=\lceil{b-a}\rceil.\
 \end{align*}
But $\lfloor{b-a}\rfloor=\lceil{b-a}\rceil$, implies that
$b-a=\lim\limits_{N\rightarrow\infty}{\frac{1}{N}\textit{card}\left(I\bigcap\frac{1}{N}\mathbb{Z}\right)}$.
Thus, it has indeed been established that $|I|=\lim\limits_{N\rightarrow\infty}{\frac{1}{N}\textit{card}\left(I\bigcap\frac{1}{N}\mathbb{Z}\right)}$.
