- Determine whether the following statements are true and false. If true, provide a proof. If false, prove a counter example.
If \(f ( x )\) is a increasing, continuous function on the interval \([ 0 , 1 ]\) such that \(f ( 0 ) = 0\) and \(f ( 1 ) = 1\) , then there exists a set \(E \subset [ 0 , 1 ]\) of positive measure such that \(f ^ { \prime } ( x ) > 0\)
If \(f ( x )\) is a strictly increasing, absolutely continuous function on the interval [0, 1] with \(f ( 0 ) = 0\) and \(f ( 1 ) = 1\) , then the set \(f ^ { - 1 } ( E ) \cap \{ x \in [ 0 , 1 ] : f ^ { \prime } ( x ) > 0 \}\) is measurable for any measurable set \(E \subset [ 0 , 1 ]\)
Solution
1. Part (a) is false.
Take \(f\) to be the Cantor–Lebesgue function. Then \(f:[0,1]\to[0,1]\) is continuous and increasing, with \[ f(0)=0, \qquad f(1)=1. \] However, \[ f'(x)=0 \] for almost every \(x\in[0,1]\). Hence \[ m\{x:f'(x)>0\}=0. \] So there need not exist a positive-measure set on which \(f'>0\).
2. Reduce part (b) to the preimage of a null set.
Now assume \(f\) is strictly increasing and absolutely continuous, with \(f(0)=0\) and \(f(1)=1\). Then \(f\) is a homeomorphism of \([0,1]\) onto itself.
Let \[ D:=\{x\in[0,1]:f'(x)>0\}, \] where \(f'\) is defined arbitrarily on the null set where the derivative does not exist. Since derivatives of absolutely continuous functions are measurable, \(D\) is measurable.
Let \(E\subset[0,1]\) be Lebesgue measurable. Write \[ E=B\cup N, \] where \(B\) is Borel and \(N\) is null. Then \[ f^{-1}(E)\cap D =\bigl(f^{-1}(B)\cap D\bigr) \cup \bigl(f^{-1}(N)\cap D\bigr). \] The first set is measurable because \(f\) is continuous. It remains to control the second.
3. Show that the null part has null preimage on \(D\).
Set \[ A:=f^{-1}(N)\cap D. \] Because \(f\) is absolutely continuous and strictly increasing, the change-of-variables formula for monotone absolutely continuous functions gives, for every measurable \(A\subset[0,1]\), \[ m(f(A))=\int_A f'(x)\,dx. \] Since \[ f(A)\subseteq N, \] we have \(m(f(A))=0\). Therefore \[ \int_A f'(x)\,dx=0. \] But \(f'(x)>0\) on \(A\). Hence \(m(A)=0\).
Thus \(f^{-1}(N)\cap D\) is measurable, and therefore so is \[ \boxed{f^{-1}(E)\cap\{f'>0\}.} \] So part (b) is true.