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The following is Exercise 7:5.3 from Bruckner's Real Analysis:

Apply Theorem 7.22 to an appropriately chosen function $f$ to prove that there exists an absolutely continuous function $F$ that is nowhere monotonic. That is,for every $c,d \in \mathbb{R}$ such that $a≤c<d≤b$, $F$ is not monotonic on $[c, d]$.

Theorem 7.22 Let $f$ be Lebesgue integrable on $[a, b]$, and let $F(x)= \int_a^x f dλ$ for $x \in [a,b]$. Then $F$ is differentiable at almost every point, and $F'=f$ almost everywhere.

How can I apply Theorem 7.22 to do the exercise? Also there seem to be not an easy function to construct (Section 4). And how part (b) of this question is absolutely continuous and how it is nowhere monotone?

  • So have you tried to find the function $f$ such that $$\lambda( { f>0}\cap[c,d] )>0$$ and $$\lambda( { f<0}\cap[c,d] )>0$$ for all nontrivial interval $[c,d] \subset [a,b]$? – Paresseux Nguyen Jun 03 '21 at 15:56
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    @ParesseuxNguyen, that was exactly my thought too! But no function came to my mind being AC –  Jun 03 '21 at 15:58
  • Why you need $f$ to be AC? Isn't it $F$ that has to be AC? – Paresseux Nguyen Jun 03 '21 at 15:59
  • Anyways, when you finish your proof ( I don't think this question is hard for you if you have the idea), it would be nice to post it on MSE. It would be a good reference for later as I didn't find any question same as yours on MSE. – Paresseux Nguyen Jun 03 '21 at 16:10

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Hint:

  • Consider a Borel set $E$ such that for any open interval $I\subset \mathbb{R}$, $0<\lambda(E\cap I)<\lambda(I)$. You might've constructed such set $E'$ before for in $[0,1]$ with the desired property for intervals $I\subset[0,1]$. The set $E=\bigcup_{n\in\mathbb{Z}}(E'+n)$ will do.

  • Notice that $E^c:=\mathbb{R}\setminus E$ has a similar property. Indeed,
    $$\lambda(I)=\lambda(I\cap E)+\lambda(I\cap E^c)$$ and since $0<\lambda(E\cap I)<\lambda(I)$, we obtain that $0<\lambda(I\cap E^c)<\lambda(I)$.

  • Define $F(x)=\int^x_0\Big(\mathbb{1}_E-\mathbb{1}_{E^c}\Big)\,d\lambda$. $F$ is absolutely continuous and $F'=\mathbb{1}_E-\mathbb{1}_{E^c}$ $\lambda$-a.s.. Since $E$ and $E^c$ are dense, no interval of positive length has fixed signed, so $F$ is not monotone in $I$.

If you have not worked on the existence of sets such as $E'$, let me know and I can sketch a construction for you.

Check if $F$ passes muster. Otherwise let me know to either fix or remove this post.

Mittens
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  • Yes, on MSE (https://math.stackexchange.com/questions/57317/construction-of-a-borel-set-with-positive-but-not-full-measure-in-each-interval) and Ex-2:13.9 of Bruckner's book! –  Jun 03 '21 at 19:29
  • @There you go. I remember it has to do with construction of fat cantor sets put together in a clever way. Anyway, now that you can look at the details about the existence of such sets $E$, see if the function $F$ that I defined pass muster. – Mittens Jun 03 '21 at 19:34
  • Why $E$ and $E^c$ are dense? How denseness of $E$ and $E^c$ results in "no interval of positive length has fixed signed"? –  Jun 03 '21 at 19:34
  • They are dense pecans they intersect every open set (in fact $\lambda(E\cap I)>0$ for all open intervals, so $E\cap I\neq\emptyset$ for all open intervals. – Mittens Jun 03 '21 at 19:55
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    If $F$ were monotone in an interval $J$, then the derivative $F'$ there would be either nonnegative a.s. in$J$ or nonpositive a.s in $J$. But or negative, $J$ contains points in $E$ and $E^c$ (with positive mass) so $\lambda({F'>0}\cap J)>0$ and $\lambda({F'<0}\cap J)>0$. $F'$ takes values $1$ and $-1$ with positive probability, that will give you a contradiction. – Mittens Jun 03 '21 at 20:00