The Cantor set is made by starting with the closed interval $[0,1]$ and removing the open middle third. $K_1=[0, 1/3]\cup [2/3, 1]$, $K_2=[0, 1/9]\cup [2/9, 1/3]\cup [2/3, 7/9]\cup [8/9, 1]$. Continue removing the middle thirds to form $K_3, K_4, \dots$. The Cantor set $K$ is the limit or intersection of these sets $K=\bigcap _{n=1}^\infty K_n$. Show that $K$ is a Borel set and find its length or Lebesgue measure.
Not understanding this question. My book defined Borel sets to be rectangles, unless I am misinterpreting the meaning of these two sentences?
The Lebesgue measure of a set $A$ can be defined for any set $A$ in a $\sigma$-field $\mathcal A$ called the Borel sets of $\mathbb R^n$. Formally, $\mathcal A$ is the smallest $\sigma$-field that contains all "rectangles" $(a_1, b_1)\times \dots \times (a_n b_n)=\{x\in\mathbb R^n:a_i<x_i<b_i, i=1,\dots,n\}$
Since the Cantor set is made up of infinitely many points it couldn't possibly be a rectangle (open interval)? And how would you calculate the Lebesgue measure? I want to say $0$ bc it's less than $2/3$ and less than $4/9$ etc.
The Lebesgue measure of a Cantor set is $0$. By definition, Cantor set is uncountable, yet the measure is $0$. To show this, you need to consider the definition of the Cantor set $\mathcal{C}$ as a limit of a decreasing sequence. As you have described, Cantor set is constructed using closed sets. Recall that closed sets are in the Borel $\sigma$-algebra (which by the way is the smallest $\sigma$-algebra containing closed sets).
$\sigma$-algebras are closed under union, and hence if you union all $K_n$'s, the union must be measurable as well. Now, you write the Cantor set as
$\mathcal{C}=\bigcap_{n\in \mathbb{N}}K_n$
and it follows from the fact that $\sigma$-algebras are closed under countable intersection that $\mathcal{C}$ (or $K$ as you call it) must be measurable. Now, if you let $\mu$ be a Lebesgue measure, you have
$\mu(\mathcal{C})=\lim_{n\rightarrow \infty}(K_n)$
where
$\mu(K_n)=\Big(\frac{2}{3} \Big)^n$ for all $n\in \mathbb{N}$.
If you take the limit of $\mu(K_n)$, what do you get?