A question concerning the Cantor (ternary) function I came across the following problem in my self-study:
Let $f:[0,1] \rightarrow [0,1]$ be the Cantor (ternary) function, and let $g(x) =  f(x) + x$. Then:
(a) $g$ is bijection from $[0,1]$ to $[0,2]$, and $h = g^{-1}$ is continuous. 
(b) If $C$ is the Cantor (ternary) set, $m(g(C))=1$.
(c) It follows that $g(C)$ contains a Lebesgue nonmeasurable set $A$. Let $B = g^{-1}(A)$, and show that $B$ is Lebesgue measurable but not Borel.
(d) There exist a Lebesgue measurable function $F$ and a continuous function $G$ on $\mathbb{R}$ such that $F \circ G$ is not Lebesgue measurable.
I think I can prove (a), but I am not having any nice ideas on how to proceed for the remaining (3) parts, and I am interested to see if anyone visiting knows how to tackle this interesting exercise. 
 A: For (b), the key observation is that, as $g$ is continuous, the image of an interval is an interval, and so $m(g((a,b)))=g(b)-g(a)$. 
Then, as Tim said, we can use that the complement of $C$ is a countable disjoint union of open intervals  $C_k$ whose measures add to 1, and that on these intervals we have $f=0$. So, if $C_k=(a_k,b_k)$, then 
$$
m(g(C_k))=g(b_k)-g(a_k)=f(b_k)-f(a_k)+b_k-a_k=b_k-a_k=m(C_k).
$$
So
\begin{align}
2-m(g(C))&=m([0,2]\setminus g(C))=m(g(\bigcup_k C_k))=m(\bigcup_k g(C_k))\\ \ \\ &=\sum_k m(g(C_k))
=\sum_k m(C_k)=1,
\end{align}
i.e. $m(g(C))=1$.
For (c), it is an established result that any set of positive measure contains a non-measurable subset. So let $A\subset g(C)$ be non-measurable, and put $B=g^{-1}(A)$. Since $B\subset C$ and $C$ is a null-set, $B$ is also a null-set, and it particular it is measurable. But it is not Borel: since $g^{-1}$ is continuous, the pre-image of a Borel set is Borel; as $A=g(B)$, this would make $A$ Borel and thus measurable, which it's not. 
Finally, for (d), the key is that
$$
1_A=1_{g(B)}=1_B\circ g^{-1}.
$$
The characteristic $1_A$ is not measurable since $A$ isn't. But $B$ is measurable so $1_B$ is, and $g^{-1}$ is continuous. So the composition of measurable functions may fail to be measurable, even if one of them is continuous. 
