# $f$ is an increasing function, absolutely continuous such that $f'=0$

i. Give an example of a function strictly increasing $$f$$ and absolutely continuous on $$[0,1]$$ such that $$f'=0$$ into some set of positive measure.

ii. Show that there exists a set of measure zero $$E \subset [0,1]$$ such that $$f^{-1}(E)$$ is not measurable.

My attempt so far:

Let $$A$$ be the complement of the generalized Cantor's set $$C_\alpha$$. We know that $$m(C_\alpha)=1-\alpha>0$$ and taking $$f$$ as the indefinite integral of $$\chi_A$$ we have that $$f$$ is absolutely continuous. But I don't know how to verify the rest of the conditions i and ii. Any hint?

• What does it mean to say '$f'=0$ into some set of positive measure'? Jun 22, 2021 at 5:30
• For (ii), the distinction between Lebesgue measurable and Borel measurable really matters. You may need to clarify. Jun 22, 2021 at 7:16
• Is $C_{\alpha}$ any Cantor set (i.e. a nonempty perfect nowhere dense set) with measure $1-\alpha,$ or is $C_{\alpha}$ further restricted in some way, such as being constructed by a dyadic process in which centrally located open intervals are removed? I think it doesn't matter, but it's probably best to say explicitly which. If further restricted, you don't need to give all the tedious details of construction. Saying symmetric Cantor set with measure $1-\alpha$ (or whatever type you have in mind) is probably enough. Jun 22, 2021 at 8:40
• I posted a lot of relevant literature here, in case you or others are interested. Jun 22, 2021 at 8:50

Part (i):

Let $$A$$ be the complement of the generalized Cantor set $$C_{\alpha}$$. Let $$f:[0,1]\rightarrow\mathbb{R}$$ be defined by $$f(x)=\int_{0}^{x}\chi_{A}(t)dt$$. From Lebesgue integration theory, $$f$$ is absolutely continuous and $$f'=\chi_{A}$$ a.e. Let $$B=\{x\in[0,1]\mid f'(x)\mbox{ exists}\mbox{ and }f'(x)=\chi_{A}(x)\}$$. Note that $$0. However, for $$x\in C_{\alpha}\cap B$$, we have that $$f'(x)=\chi_{A}(x)=0$$. Therefore, $$f'=0$$ on a set of positive measure.

Next, we show that $$f$$ is strictly increasing. Prove by contradiction. Suppose that there exist $$x_{1},x_{2}\in[0,1]$$ such that $$x_{1} and $$f(x_{1})=f(x_{2})$$. Therefore, $$\int_{x_{1}}^{x_{2}}\chi_{A}(t)dt=0$$ which implies that $$m((x_{1},x_{2})\cap A)=0$$. Note that $$(x_{1},x_{2})\cap A$$ is an open subset of $$\mathbb{R}$$. It has zero Lebesgue measure $$\Rightarrow$$ $$(x_{1},x_{2})\cap A=\emptyset$$. It follows that $$(x_{1},x_{2})\subseteq C_{\alpha}$$, which is a contradiction because generalized Cantor $$C_{\alpha}$$ does not have interior.

Part (ii):

Clearly $$f:[0,1]\rightarrow[0,\alpha]$$ is a strictly increasing bijection. (Note that $$f(1)=m(A)=\alpha$$.)

Claim: $$m(f(B))=\int_{B}f'$$ for any Borel subset $$B$$ of $$[0,1]$$.

Proof of Claim: Let $$\mathcal{P}=\{[0,x]\mid x\in[0,1]$$}. Let $$\mathcal{L}$$ be the collection of all Borel subsets $$B$$ of $$[0,1]$$ such that $$m(f(B))=\int_{B}f'$$. We verify that $$\mathcal{P}\subseteq\mathcal{L}$$, $$\mathcal{P}$$ is a $$\pi$$-class, and $$\mathcal{L}$$ is a $$\lambda$$-class (with respect to $$[0,1]$$), then we invoke Dynkin's $$\pi$$-$$\lambda$$ Theorem.

Clearly, for any $$B_{1},B_{2}\in\mathcal{P}$$, $$B_{1}\cap B_{2}\in\mathcal{P}$$. Let $$g:[0,\alpha]\rightarrow[0,1]$$ be defined by $$g=f^{-1}$$. Clearly $$g$$ is strictly increasing and hence Borel. If $$B\subseteq[0,1]$$ is Borel, then $$f(B)=g^{-1}(B)$$ is a Borel subset of $$[0,\alpha]$$ and hence $$m(f(B))$$ is well-defined. Obviously $$B\mapsto m(f(B))$$ is $$\sigma$$-additive, i.e., it is a measure. Denote the finite measures $$B\mapsto m(f(B))$$ and $$B\mapsto\int_{B}f'$$ by $$\mu$$ and $$\nu$$ respectively. Clearly $$\mu([0,1])=m(f([0,1]))=m([0,\alpha])=\alpha$$ and $$\nu([0,1])=\int_{[0,1]}\chi_{A}(t)dt=m(A)=\alpha$$. It follows that $$[0,1]\in\mathcal{L}$$. It is routine to verify that $$B\in\mathcal{L}\Rightarrow[0,1]\setminus B\in\mathcal{L}$$ and $$\cup_{n}B_{n}\in\mathcal{L}$$ whenever $$B_{1},B_{2},\ldots\in\mathcal{L}$$ are pairwisely disjoint. Therefore $$\mathcal{L}$$ is a $$\lambda$$-class. Finally, if $$x\in[0,1]$$, then $$\mu([0,x])=m([0,f(x)])=f(x)$$ while $$\nu([0,x])=\int_{[0,x]}f'(t)dt=f(x)-f(0)=f(x).$$ Hence, $$\mathcal{P}\subseteq\mathcal{L}$$. By Dynkin's theorem, $$\sigma(\mathcal{P})\subseteq\mathcal{L}$$, which implies that $$\mathcal{L}=\mathcal{B}([0,1])$$ because $$\sigma(\mathcal{P})=\mathcal{B}([0,1])$$. That is, $$m(f(B))=\int_{B}f'$$ for any Borel subset $$B$$ of $$[0,1]$$.

Now, we go back to your question. By Part(i), there exists a Lebesgue measurable set $$B_{1}$$ with $$m(B_{1})>0$$ such that $$f'=0$$ on $$B_{1}$$. Recall that the $$\sigma$$-algebra of Lebesgue measurable sets is just the completion of the $$\sigma$$-algebra of Borel measurable sets, so we can choose a Borel set $$B_{2}$$ such that $$m(B_{2}\Delta B_1)=0$$. Then, $$m(B_2)=m(B_1)>0$$ and $$f'=0$$ a.e. on $$B_2$$. Recall the fact that every Borel set that has a positive measure contains a non-Lebesgue measurable set. Choose a non-measurable set $$D\subseteq B_{2}$$. Note that $$f(D)\subseteq f(B_{2})$$ and $$m(f(B_{2}))=\int_{B_{2}}f'=0$$ because $$f'=0$$ a.e. on $$B_{2}$$. Therefore $$f(D)$$ is Lebesgue measurable by the completeness of Lebesgue measure. Moreover, $$m(f(D))=0$$. Finally, $$f^{-1}(f(D))=D$$ is non-Lebesgue measurable.

• I liked the answer. It is very well explained. Thanks Jun 22, 2021 at 18:25