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.
