Prove the uniformity of the Cantor/Lebesgue function defined on $A^c$ where $A$ is a Cantor set on $[0,1]$

Question

I am reading Lebesgue Integration on Euclidean Space by Frank Jones. My question is specifically regarding Chapter 4, Section C titled "The Lebesgue Function Associated with a Cantor Set". The author denotes the function as $f$ and defines it on the union of the open intervals $J_r$ of a Cantor set in $[0,1]$.

On page 87, the author picks two points $x,y \in A^c$ and $|x-y| \leq \mathcal{l}_3$ where $\mathcal{l}_3$ is the length of the closed intervals of the Cantor set. Of course, if neither $x$ nor $y$ belongs to one of these closed intervals, then $x$ and $y$ must belong to the same excluded open interval so $f(x)=f(y)$. This makes sense to me.

However, the author then goes on to state,

We therefore assume that either $x$ or $y$ (or both) belongs to one of these closed intervals.

This above statement confuses me If we're trying to prove something for a function defined on $A^c$, which are the open intervals removed from the Cantor set, then why/how can we assume that $x$ or $y$ belongs to one of the closed intervals?

It would help if someone that has the book could opine, although I hope my explanation is clear.

Answer 1

You should keep in mind the difference between the Cantor set $A$ (which does not contain any intervals), and "pre-Cantor" sets, which consist of $2^k$ disjoint closed intervals of equal length $l_k$. It seems the author does not introduce any notation for pre-Cantor sets; I'll call them $A_k$. So, $$A = \bigcap_{k=1}^\infty A_k$$ The function $f$ is defined on the complement of $A$. The author estimates $|f(x)-f(y)|$ from above, as follows:

1. Pick $x,y\in A^c$
2. Suppose there is $k$ such that $|x-y|\le l_k$
3. Then either
   • $x,y\in A_k^c$, in which case $f(x)=f(y)$, or
   • $x\in A_k$, $y\in A_k^c$, or
   • $x\in A_k^c$, $y\in A_k$ (which is essentially the same as preceding), or
   • $x\in A_k$, $y\in A_k$

The author then proceeds to estimate $|f(x)-f(y)|$ considering each case separately.