If $f$ is absolutely continuous and monotonic on a compact interval, then the flat part of $f$ consists of at most a countable number of segments.

Question

I have a question that is seemingly somewhat similar to Froda's theorem.

Suppose $f : [0,1] \rightarrow [0,1]$ is absolutely continuous and monotonic with $f(0)=0$ & $f(1)=1$, and let $\Omega$ be the union of maximal segments (i.e. open intervals) on which $f$ is constant (if there are any). Is it true that the number of such (disjoint) segments is at most countable? I must know the answer to this question in order to resolve some measurability issue. Thanks a lot!

(add : my understanding from the first answer) : Now, let $\Sigma$ be the union of the maximal half-closed intervals (i.e. intervals of the form $(a_i, b_i]$, instead of segments of the form $(a_i,b_i)$) on which $f$ is constant. Then, the restriction $g$ of $f$ to $[0,1]/\Sigma$ is bijective so that the inverse $g^{-1} : [0,1] \rightarrow [0,1]/\Sigma$ is well-defined and bijective as well. (Note that I used the half-closed intervals instead of the segments in order to make the restriction g be injective.) Now, since $g^{-1}$ is monotonic on the compact interval $[0,1],$ we obtain from Froda's theorem that the discontinuities of $g^{-1}$ are at most countable, implying that $\Sigma $ is a countable union of half-closed intervals, which in turn yields the desired result.

Answer 1

Another answer speaks of a "basis" for a topology. But we don't need to know about that here.

A set of pairwise disjoint open intervals in $\mathbb R$ must be at most countably infinite because every open interval contains rational numbers (the proof of that can be an interesting exercise: between any two reals there is a rational). And there are only countably many rationals.

Answer 2

Yes, this is true. And as you pointed out, it is very similar to Froda's Theorem. Actually, you can use the theorem to prove your statement.

Using Froda, you know that the number of discontinuities is countable. Now, if the number of constant intervals were uncountable, you would need to have an uncountable number os discontinuities, since every constant interval would have a different value (due to monotonicity), which is a contradiction. Hence, you have countable constant intervals.

Answer 3

If $\Omega \subset [0, 1]$ is any union of open intervals, then $\Omega$ can always be written as a countable union of open intervals. This is because $[0, 1]$ is second-countable: the open intervals with rational endpoints form a basis for the topology on $[0, 1]$.