# Prove that $f' = 0$ a.e.

I am trying to show the following statement: Let $$f: (a,b) \to \mathbb{R}$$ be increasing and $$E \subset (a,b)$$ measurable. For $$\epsilon > 0$$, there exists $$(a_i,b_i)$$ such that $$E \subset \cup(a_i,b_i), \sum_{i} f(b_i) - f(a_i) < \epsilon$$. Prove that $$f'(x) = 0$$ a.e. on $$E$$.

My thoughts: Suppose $$f$$ is absolutely continuous on $$(a,b)$$. Then $$f$$ is differentiable a.e. on $$(a, b)$$, its derivative $$f'$$ is integrable over $$(a, b)$$, and $$\int_{a_i}^{b_i} f = f(b_i) - f(a_i)$$ for all $$i$$. Then I want to show that $$\int_{a_i}^{b_i} f' = 0$$ for all $$(a_i,b_i) \subset (a,b)$$, so $$f' = 0$$ a.e.

I am stuck on two points. 1) Is $$f$$ absolutely continuous from the given condition? The definition requires a finite disjoint collection, but here I am only given a countable one. 2) I want to show that $$\int_{a_i}^{b_i} f' = 0$$ for all $$(a_i,b_i) \subset (a,b)$$, but from what's given I only know that for all $$\epsilon$$, there exists some $$(a_i,b_i)$$ such that $$\int_{a_i}^{b_i} f' < \epsilon$$, which is not equivalent to what I want to show.

Can someone give me a hint? Any help is appreciated. Thanks in advance!

There is no need to assume the absolute continuity of $$f$$. Because $$f$$ is increasing, we know that $$f'$$ exsits almost everywhere and $$\int_c^df'(x)\,dx\leq f(d)-f(c)$$ for all $$[c,d]\subset (a,b)$$; also $$f'\geq0$$ almost everywhere.
Hence, for every $$\epsilon>0$$, we have $$0\leq\int_Ef'(x)\,dx\leq\int_{\cup(a_i,b_i)}f'(x)\,dx\leq\sum_i\int_{a_i}^{b_i}f'(x)\,dx\leq \sum_i\left(f(b_i)-f(a_i)\right)<\epsilon.$$ Since $$\epsilon>0$$ is arbitrary, we have $$\int_Ef'(x)\,dx=0$$. Recalling that $$f'\geq0$$ almost everywhere, we have $$f'=0$$ a.e. on $$E$$.
Remark. Here we can't get the absolute continuity of $$f$$ on $$(a,b)$$ from the hypothesis. Whether $$f$$ is absolutely continuous depends on almost all information of $$f$$ on the whole interval $$(a,b)$$; however, the hypothesis here focus on a subset $$E$$ of $$(a,b)$$, so, of course we can't say anything about the absolute continuity of $$f$$ on $$(a,b)$$. Then you may think: What about the absolute continuity of $$f$$ on $$E$$? Well, generally we only talk about the absolute continuity of $$f$$ on some interval; but $$E$$ is not necessarily an interval. Even if $$E$$ is an interval, the absolute continuity of $$f$$ on $$E$$ is not obvious, as far as I'm concerned.