You can construct $f$ as follows: let $m$ denote Lebesgue measure. Once $m(A)=0$, we can choose for each $i=1,2,\cdots$, a sequence of intervals $I_{ij}$ such that $$A\subset \cup_{j=1}^\infty I_{ij},\ \sum_{j=1}^\infty m(I_{ij})\le \frac{1}{2^i}. \tag{1}$$
Note that each $x$ belongs to infinitely many intervals $I_{ij}$. Define $f:[0,1]\to\mathbb{R}$ by $$f(x)=\sum _{i,j=1}^\infty m(I_{ij}\cap [0,x]).$$
Note that $f$ is increasing. Also $f$ satisfies $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\infty,\ \forall x\in A.$$
Indeed, as $x$ belongs to infinitely many intervals $I_{ij}$, we can assume without of generality that $x$ belongs to $J_k=(x-a_k,x+a_k)$ for all $k=1,2,\cdots$, that each $J_k=I_{ij}$ for some $i,j$ and $a_k\to 0$. Note that
\begin{eqnarray}
\frac{f(x-h)-f(x)}{h} &=& \frac{1}{h}\sum _{i,j=1}^\infty m(I_{ij}\cap [x-h,x]) \nonumber \\
&\ge& \frac{1}{h}\sum _{k=1}^\infty m(J_k\cap [x-h,x]) \nonumber \\
&\ge& \frac{1}{h}(Nh+\sum _{k=N+1}^\infty a_k),
\end{eqnarray}
where $N$ depending on $h$ is choosen in such a way that $h\le a_N$. As $h\to 0^+$, we see from the previous inequality that the left derivative is infinity. With an similar reasoning, you can conclude that the right derivative is also infinity.
To prove that $f$ is absolutely continuous, note that $$\sum _{k=1}^N |f(x_{k+1})-f(x_k)|=\sum _{k=1}^n \sum_{i,j=1}^\infty m(I_{ij}\cap[x_k,x_{k+1}]). \tag{2}$$
Once $(1)$ is satisfied, you conclude from $(2)$ that $f$ is absolutely continuous. To get a strictly increasing functions, just consider $g(x)=x+f(x)$.
Note: The original construction is due to professor Porter and it is contained here.
Aside. Your item 3 sets an unattainable goal. If a continuous function $f$ satisfies $f'=0$ for all but countably many points of $\mathbb R$, then $f$ is a constant function. See Set of zeroes of the derivative of a pathological function.
Here is a more explicit construction than in the post link in comments. Fix $r\in (0,1/2)$. Let $f_0(x)=x$. Inductively define $f_{n+1}$ so that
- $f_{n+1} (x) = f_n(x)$ when $x\in 2^{-n}\mathbb Z$
- If $x\in 2^{-n}\mathbb Z$, let $f_{n+1}(x+2^{-n-1}) = r f_n(x) + (1-r) f_n(x+2^{-n})$.
- Now that $f_{n+1}$ has been defined on $2^{-n-1}\mathbb Z$, extend it to $\mathbb R$ by linear interpolation.
- Let $f=\lim f_n$.
Here is a piece of this function with $r=1/4$:
and for clarity, here is the family $f_0$,$f_1$,... shown together. All the construction does is replace each line segment of previous step with two; sort of like von Koch snowflake, but less pointy. (It's a monotonic version of the blancmange curve).
To check that the properties hold, you can proceed as follows:
- Since $\sup|f_{n+1}-f_{n}| \le 2^{-n}$, it follows that $f_n$ is a uniformly convergent sequence. So the limit is continuous.
- The values of $f$ at every dyadic grid $2^{-n} \mathbb Z$ are strictly increasing, by direct inspection. Since $f$ eventually stabilizes at dyadic rationals, it is strictly increasing.
- Being increasing, $f$ is differentiable almost everywhere. Let $x$ be a point of differentiability (note that $x$ is not a dyadic rational). Then
$$f'(x) = \lim_{n\to\infty} 2^{n} (f_n(x_n+2^{-n})-f_n(x_n)) \tag{1}$$
where $x_n\in 2^{-n}\mathbb Z$ is such that $x\in (x_n,x_n+2^{-n})$. The expression inside the limit in (1) is a product of the form $r^{i}(1-r)^j$ where $i+j=n$. More precisely, $i$ and $j$ are the numbers of $1$s and $0$s in the first $n$ digits of the binary expansion of the fractional part of $x$. It follows that $f'(x)=0$ unless $x$ has a lot more $0$s than $1$s in the binary expansion; but the number of such points $x$ has zero measure.
(It may be more enjoyable to prove part 3 using the Law of Large Numbers from probability.)
Best Answer
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 $0<m(C_{\alpha})=m(C_{\alpha}\cap B)+m(C_{\alpha}\cap B^{c})=m(C_{\alpha}\cap B)$. 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 $x_{1}<x_{2}$ 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.