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suppose $f$ is an absolutely continuous on $[0,1]$,that almost everywhere $f'>0$.

is the inverse of $f$ necessarily absolutely continuous on $[f(0),f(1)]$?

thank you very much!

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It is always useful to have more than one answer to a problem. One learns quite a bit more.

Even better, however, is to have two contradictory answers to the same problem. That is both more entertaining and more educational. (The contradiction is just about what one wants to prove, not about errors.)

A friend of mine often tells an anecdote about a similar situation. A well-known mathematician had posed an open problem at the end of one of his papers. My friend sent him a solution. He wrote back quite pleased and informed my friend that someone else had also submitted a solution using a completely different method. He proposed that both should be published back-to-back in the Revue Roumaine to which, as an editor, he would submit them. And they were published. The methods were indeed completely different. My friend had solved the problem positively and the adjacent paper "proved" the opposite.

Problem. Suppose that $f:[0,1]\to\mathbb{R}$ is absolutely continuous and that $f'(x)>0$ for a.e. $x\in [0,1]$. Prove that $f$ has an absolutely continuous inverse.

Proof. Clearly $f$ is continuous and strictly increasing on $[0,1]$ and so it has a continuous and strictly increasing inverse $f^{-1}$. Let $P$ be the set of points at which $f$ has a finite, positive derivative. Let $Z$ be the remaining points which we know is a set of measure zero. Since $f$ is AC it satisfies Lusin's condition (N) so $f(Z)$ is also a set of measure zero. Let $E_1= f(P)$ and $E_2=f(Z)$. These sets exhaust $[f(0),f(1)]$. We know that $f^{-1}$ has a finite positive derivative at each point of $E_1$ and that $E_2$ has measure zero. The function $f^{-1}$ maps any measure zero subset of $E_1$ to a set of measure zero (because it has a finite derivative there). Also $f^{-1}$ maps every subset of $E_2$ to a set of measure zero (i.e., a subset of $Z$). Thus $f^{-1}$ satisifies Lusin's condition (N) on $[f(0),f(1)]$. It follows that $f^{-1}$ is absolutely continuous.

[Note added: The OP remembered a proof of this statement:

"Suppose $f'(x)$ exists at each point $x\in E$ and that $|f'(x)|\leq M$ there. Then $m(f(E))\leq Mm(E)$. Hence $f(E)$ is measure zero if $E$ is measure zero.

But this can be pushed. If $E$ has measure zero and $f'(x)$ is finite at every point of $E$ ($|f'(x)|$ not necessarily bounded) then simply write $E_n=\{x\in E: |f'(x)|\leq n\}$ and use the fact that $$m(f(E)) \leq \sum_{n=1}^\infty m(f(E_n))\leq \sum_{n=1}^\infty nm( E_n)=0.$$

In fact, with bit more work, one can prove that $$m(f(D)) \leq \int_D |f'(x)|\,dx$$ for any measurable set $D$ assuming that $f$ has a finite derivative at every point of $D$.

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  • $\begingroup$ note that "The function f−1 maps any measure zero subset of E1E1 to a set of measure zero (because it has a finite derivative there)." Is not necessarily true,because you must prove also that on that measure zero set ,derivative has a max. $\endgroup$ – user115608 Nov 25 '15 at 6:39
  • $\begingroup$ @user115608 Not true. I know why you might think that, but it is true that any function that has a derivative at every point of a set does indeed satisfy the Lusin (N) condition there. Think about why you want a bounded derivative and then consider splitting the set into pieces on which you do have a bounded derivative and then adding them up. $\endgroup$ – B. S. Thomson Nov 25 '15 at 16:01
  • $\begingroup$ so I can't believe! Your answer vs Umberto ' s!Which one is true? $\endgroup$ – user115608 Nov 25 '15 at 18:56
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    $\begingroup$ Umberto has proved that there is a strictly increasing Lipschitz function on $[0,2]$ whose inverse is not absolutely continuous. He didn't promise that it had a positive derivative almost everywhere. (It doesn't). If you add in that hypothesis then the counterexample disappears. You should include both in your answer as the Umberto example shows that the hypothesis cannot be weakened to: "$f$ is absolutely continuous and strictly increasing." (Not even to: "$f$ is Lipschitz and strictly increasing".) $\endgroup$ – B. S. Thomson Nov 25 '15 at 19:00
  • $\begingroup$ yes.u r right! Thank you very much.would you please see this?math.stackexchange.com/questions/1546261/… $\endgroup$ – user115608 Nov 25 '15 at 19:02
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Let $c: [0,1] \to [0,1]$ be the usual Cantor-Lebesgue function and let $g: [0,1] \to [0,2]$ be given by $g(x) = c(x) + x$. Then $g$ is continuous, increasing, and $g'(x) = 1$ almost everywhere. Since $$g(1) - g(0) > \int_0^1 g'(x) \, dx$$ it follows that $g$ is not absolutely continuous.

Let $f = g^{-1}(t)$ so that $f : [0,2] \to [0,1]$ is continuous and increasing. If $x,y \in [0,2]$ and $x < y$ then $$\left| \frac{y-x}{f(y) - f(x)} \right| = \frac{y-x}{f(y) - f(x)} = \frac{g(f(y)) - g(f(x))}{f(y) - f(x)} = \frac{c(f(y)) - c(f(x))}{f(y) - f(x)} + \frac{f(y) - f(x)}{f(y) - f(x)}.$$ Since $c$ is nondecreasing, the last expression is greater than or equal to $1$. Thus $$|f(x) - f(y)| \le |x-y|.$$

It follows that $f$ is Lipschitz, hence AC, but $g = f^{-1}$ is not AC.

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  • $\begingroup$ your $f$ is not defined on $[0,1]$,what will happen if you consider the restriction? $\endgroup$ – user115608 Nov 24 '15 at 17:59
  • $\begingroup$ How about using $f_1(x) = f(2x)$? $\endgroup$ – Umberto P. Nov 24 '15 at 19:17
  • $\begingroup$ yes u r right.your beautiful counter example works.thank u.but we always say that with this assumption f is strictly increasing.true? $\endgroup$ – user115608 Nov 24 '15 at 19:25
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    $\begingroup$ I upvoted this answer because it is so lucidly and elegantly expressed and does, indeed, contribute to an understanding of the problem. It might be noted too that the function $g$ here has an infinite derivative on some set and that set maps by $g$ to a set of positive measure (at the points of which $f'$ is equal to zero). $\endgroup$ – B. S. Thomson Nov 25 '15 at 3:29

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