Suppose $f:[0,1]\to\mathbb{R}$ is a nondecreasing function, and $f'(x)=0$ almost everywhere on $[0,1]$.
For any $\epsilon>0$, prove that there exists finitely many pairwise disjoint intervals $[a_k,b_k]$, $k=1,\dots, n$ in $[0,1]$ with $$\sum_{k=1}^n(b_k-a_k)<\epsilon$$ and $$\sum_{k=1}^n(f(b_k)-f(a_k))>f(1)-f(0)-\epsilon.$$
My attempt: I think proving by contradiction may be the way to go here.
Suppose to the contrary there exists $\epsilon>0$ such that for any finitely many pairwise disjoint $[a_k,b_k]\subseteq[0,1]$ with $\sum_{k=1}^n(b_k-a_k)<\epsilon$, we have $$0\leq\sum_{k=1}^n(f(b_k)-f(a_k))\leq f(1)-f(0)-\epsilon.$$
This looks like a weaker condition than absolutely continuous.
However, I am not sure how to proceed here, other than noting that $f(1)-f(0)>0$, so that $f$ is not constant. Also, I can see that $f$ cannot be absolutely continuous, otherwise it will have a contradiction, since an absolutely continuous and singular function must be constant.
Thanks for any help!
Update:
I have another idea, let $E$ be the set where $f'\neq 0$. Then $|E|=0$.
My idea is to cover $E$ in the Vitali sense by intervals $[a_k,b_k]$ and use Vitali Covering Theorem to prove that $\sum(b_k-a_k)<\epsilon$, but $[a_k, b_k]$ covers nearly all of $E$, where all the changes occurs, so that $\sum(f(b_k)-f(a_k))>f(1)-f(0)-\epsilon$.
At the moment, I have no idea how to make the above idea rigorous though.