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!


1 Answer 1


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.

The above is a hint. Detailed proof is below.

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.

  • $\begingroup$ I see! Thank you so much for your insightful answer! $\endgroup$ Oct 21, 2022 at 6:54
  • $\begingroup$ @InsultedByMathematics You’re more than welcome. I’m very glad to help :) $\endgroup$
    – Feng
    Oct 21, 2022 at 7:18

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