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Rudin's Theorem 7.25 states that if $f:[a,b]\rightarrow R$ is differentiable everwhere in $[a,b]$ and $f^\prime \in L^1$ then $f$ is absolutely continuous (AC). In my case, I also have that $f$ is monotone.

So far I have an ugly proof--which I hope is correct--that $f^\prime \in L^1$ but I was wondering if there is some direct result stating that monotone and diff. everywhere implies AC that I can cite instead.

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    $\begingroup$ Combine his Theorem 7.21 with Theorem 6.11. $\endgroup$ – Qiyu Wen Jul 30 '16 at 3:03
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    $\begingroup$ @QiyuWen My copy of Rudin is missing. Could you remind us what 6.11 says? $\endgroup$ – David C. Ullrich Jul 30 '16 at 17:34
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    $\begingroup$ @DavidC.Ullrich It says complex measure $\lambda$ is absolutely continuous with respect to positive measure $\mu$ iff for any $\epsilon > 0$, exists $\delta>0$ such that $|\lambda(E)| <\epsilon$ for any measurable $E$ with $\mu(E)<\delta$. $\endgroup$ – Qiyu Wen Jul 30 '16 at 22:48
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    $\begingroup$ @QiyuWen Thanks. I was about to say I didn't see exactly how that settled the question, when I realized that you also mentioned 7.21; I'd misread that as 7.25. Sorry to be tedious - what's 7.21? $\endgroup$ – David C. Ullrich Jul 31 '16 at 2:08
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    $\begingroup$ @DavidC.Ullrich It says if $f:[a,b]\to \mathbb{R}$ is differentiable everywhere in $[a,b]$ and $f'\in L^1([a,b])$, then $f(x)-f(a) = \int_a^x f'\,dm$ for all $x\in [a,b]$. $\endgroup$ – Qiyu Wen Jul 31 '16 at 2:20
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On second thought, the original solution I posted is more than you need if you are willing to cite Rudin.

Assume that $f : [a,b] \to \mathbb R$ is nondecreasing. Extend $f$ to all of $\mathbb R$ by $f(x) = f(b)$ if $x > b$ and $f(x) = f(a)$ if $x < a$.

Define $f_n(x) = n[f(x+1/n) - f(x)]$. Then each $f_n \ge 0$ on $\mathbb R$ and $f_n \to f'$ almost everywhere. Fatou's lemma implies that $$ \int f' \, dx \le \liminf_n \int f_n \, dx.$$

The proof finished rather quickly since if $\frac 1n < b-a$ you get $$\int f_n \, dx = f(b) - f(a)$$ so that $f' \in L^1$.


It is well-known that $f$ is absolutely continuous on $[a,b]$ if and only if it is continuous, has bounded variation, and carries sets of measure zero to sets of measure zero.

Assuming that $f$ is differentiable everywhere in $[a,b]$ you get continuity for free, and since $f$ is monotone its variation is simply $|f(b) - f(a)|$.

It's not too hard to show using a Vitali covering argument that if $E \subset [a,b]$, $f$ is differentiable at every point of $E$, and $|f'(x)| \le k$ for all $x \in E$, then $m^*(f(E)) \le k m^*(E)$ where $m^*$ is the Lebesgue outer measure. Consequently if $f$ is differentiable at every point of $[a,b]$ and $N \subset [a,b]$, then $m^*(N) = 0$ implies $m^*(f(N)) = 0$.

This gives you AC as needed.

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  • $\begingroup$ I was just about to post your "second thought"... $\endgroup$ – David C. Ullrich Jul 29 '16 at 15:05

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