Exercise 7.3.4

By Theorem 7.3.2, $g$ is integrable. But for any rational number $q$, $f(q)>0$, and so $g\circ f$ is just Dirichlet’s function, which is not integrable.

Let $c(x)$ be the Cantor Function, defined in Exercise 6.2.12, and consider $d(x)=c(x)+x$, defined over $[0,1]$. Recall that $c(x)$ is increasing and continuous; therefore $d(x)$ is strictly increasing and continuous. Therefore, since $d$ is one-to-one, $d^{-1}$ is defined, continuous, and increasing over $[0,2]$.

Notably, $c$ maps the Cantor set $C$ (which has measure zero) to a set of positive measure, since $c$ maps $[0,1]\setminus C$ to the dyadic points in $[0,1]$, which are countable and thus have measure zero. This behaviour is also true of $d$. To prove this a bit formally, assume for contradiction that $d(C)=\{d(x):x\in C\}$ has measure zero, and find a collection of open intervals with total length less than $𝜖$ whose union contains $d(C)$. For a representative interval $(p_1,p_2)=(d(x_1),d(x_2))$, let $d(S)$ be the set of points in $d(C)$ which are covered by $(p_1,p_2)$, where $S$ be the points in $C$ which map to $d(S)$. But note that then $(c(x_1),c(x_2))\supseteq c(S)$, and that

Repeating this for every interval leaves us with a collection of intervals covering $c(C)$ whose total length is less than $𝜖$, a contradiction since $c(C)$ has positive measure.

Finally, let $f=d^{-1}$ and $g$ be the indicator function of $C$; that is,

We show later in Exercise 7.3.9 that $g$ is integrable and its set of discontinuities is $C$. But I claim that $g\circ f$ is discontinuous on $d(C)$, a set of positive measure. This claim would imply that $g\circ f$ is not integrable.

Note that $f^{-1}=d$ is continuous. If $g\circ f$ was continuous on any point $x\in d(C)$, then $g=(g\circ f)\circ f^{-1}$ should be continuous on $d^{-1}(x)\in C$, but $g$ is discontinuous over $C$. Thus a contradiction and $g\circ f$ is discontinuous over $d(C)$.