Let $\lambda$ denote Lebesgue measure and let $f: [0,1] \rightarrow [0,1]$ be a differentiable function such that for every Lebesgue measurable set $A \subset [0,1]$ one has $\lambda(f^{-1}(A)) = \lambda (A)$. Prove that either $f(x) = x$ or $f(x) = 1- x$.

I will appreciate a hint or a solution that doesn't use ergodic theory as this is an old qual problem in measure theory. $f$ is bounded and continuous and so it is integrable and hence the hypothesis of Lebesgue's Differentiation Theorem are satisfied, but I haven't seen a way to use this or if it is even applicable.

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    $\begingroup$ To understand why the "differentiable" hypothesis is there: examine the function $f(x) = |2x-1|$ on $[0,1]$. $\endgroup$ – GEdgar Apr 23 '16 at 18:15

First note $f$ is onto: Otherwise the range of $f$ is a proper subinterval $[a,b],$ which implies $\lambda(f^{-1}([a,b])) = \lambda([0,1]) = 1 > b-a = \lambda([a,b]),$ contradiction.

Suppose $f'(x) = 0$ for some $x\in [0,1].$ Then $f(B(x,r))\subset B(f(x),r/4)$ for small $r>0.$ ($B(y,r)$ denotes the open ball with center $y$ and radius $r$ relative to $[0,1].$) Thus for such $r,$ $f^{-1}(B(f(x),r/4))$ contains $B(x,r).$ But $B(f(x),r/4)$ has measure no more than $2\cdot (r/4) = r/2,$ while $B(x,r)$ has measure at least $r.$ That's a contradiction. Hence $f'(x) \ne 0$ for all $x\in [0,1].$

By Darboux, $f'>0$ on $[0,1]$ or $f'<0$ on $[0,1].$ Let's assume the first case holds. Then $f$ is strictly increasing, hence a bijection on $[0,1].$ Thus if $A$ is an interval, $\lambda(A) = \lambda(f^{-1}(f(A)))= \lambda(f(A)).$ Since $f(0) = 0,$ this gives $x = \lambda([0,x]) = \lambda(f([0,x])) = f(x) - f(0) = f(x)$ for all $x\in [0,1].$

  • $\begingroup$ This is a much better proof! $\endgroup$ – Giovanni Apr 23 '16 at 16:52

The following argument should work under the additional assumption that $f'$ is integrable. I don't know if this is something that you need to assume or if it is a limitation of my proof. It is worth mentioning that I need integrability of the derivative for two reasons: first to apply Lebesgue differentiation theorem for $f'$, and second to get that $f$ is absolutely continuous, which I need to apply the Fundamental Theorem of Calculus.

Let $\mu(\cdot) = \lambda(f^{-1}(\cdot))$. Then $\mu = \lambda$ and hence for a.e. $x \in (0,1)$ we have

$$1 = \frac{d\lambda}{d\mu}(x) = \lim_{r \to 0}\frac{\lambda(B_r(x))}{\mu(B_r(x))} = \lim_{r \to 0}\frac{\int_{B_r(x)}d\lambda}{\mu(B_r(x))} = \lim_{r \to 0}\frac{1}{\lambda(f^{-1}(B_r(x)))}\int_{f^{-1}(B_r(x))}|f'|d\lambda, $$ where in the last equality we used the change of coordinates $z = f(y)$. Notice that $\lambda(f^{-1}(B_r(x))) = \lambda(B_r(x)) \to 0$, hence by Lebesgue differentiation theorem, for every $y \in f^{-1}(x)$ we have that $|f'(y)| = 1.$ Moreover we have that for $y \in (0,1)$, $y \in f^{-1}(f(y))$, thus proving that $|f'| = 1$ a.e. in $[0,1]$.

To conclude the argument, we only need to show that $f'$ cannot change sign. Indeed, by the FTC, we would have that $$f(1) = f(0) + \int_0^1f'(y) = f(0) \pm 1,$$ showing that either $f(0) = 0$ and $f' = 1$ or $f(0) = 1$ and $f' = -1$.

To prove that the derivative is constant a.e., I'll show that $f$ is surjective. By continuity, $f([0,1])$ is a compact subset of $[0,1]$. Suppose by contradiction that $f$ is not surjective, then $\lambda(f[0,1]) < 1$. To find a contradiction it is enough to notice that $$1 = \lambda([0,1]) = \lambda(f^{-1}(f([0,1]))) = \lambda(f([0,1])) < 1.$$

In response to Ramiro's comment: Everywhere differentiable + integrable derivative implies absolute continuity, i.e. the FTC holds. Then I can prove what you asked as follows.

Necessarily, $f(0)$ is either $0$ or $1$. To see this notice that if there is $x \in (0,1)$ such that $f(x) \in \{0,1\}$ then let $y$ be the point such that $f(y) = \{0,1\} \setminus \{f(x)\}$. Assume without loss that $x< y$, then $$1 = |f(x) - f(y)| \le \int_x^y|f'| = y - x < 1.$$ Then assume without loss that $f(0) = 0$. Reasoning as above shows that $f(1) = 1$. Again by the FTC, $$1 = f(1) - f(0) = \int_0^1f' = \lambda(\{f' = 1\}) - \lambda(\{f' = -1\}).$$ Finally, if $\lambda(\{f' = -1\}) > 0$ we get a contradiction, hence $f' = 1$ a.e.

  • $\begingroup$ Why proving that $f$ is surjective implies that the derivative does not change sign? $\endgroup$ – Ramiro Apr 23 '16 at 5:12
  • $\begingroup$ @Ramiro: I have edited with a proof, let me know if it is convincing :) $\endgroup$ – Giovanni Apr 23 '16 at 5:41
  • $\begingroup$ Thanks for writing down the details. Please note that your statement "To prove that the derivative does not change sign I'll show that f is surjective." is NOT accurate. The fact that f is surjective implies only that $\lambda({f\neq 1})=0$ or $\lambda({f\neq -1})=0$. This is all you need, but it is weaker than claiming "the derivative does not change sign". (In fact, in principle, in a set of measure zero we may even have $|f'|\neq 1$). $\endgroup$ – Ramiro Apr 23 '16 at 12:48
  • $\begingroup$ Thank you for your solution. The problem mentioned that we cannot assume the derivative is continuous. I left that out because we can't in general assume things not given to us and which we cannot deduce easily as parts of the properties of the object given to us. Can we remove the assumption of integrability of the derivative or say why it suffices to consider this case only? $\endgroup$ – akech Apr 23 '16 at 13:34
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    $\begingroup$ If you allow me a suugestion, I suggest that instead of writing "To prove that the derivative does not change sign I'll show that f is surjective", it would be better to write something like "To prove that the derivative is constant a.e., I'll show that f is surjective". $\endgroup$ – Ramiro Apr 23 '16 at 16:57

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