the Riemann integrability of inverse function

If $f \colon [a,b] \rightarrow [c,d]$ is a bijection, $f\in \mathcal{R}$ and $f^{-1}$ exists, then prove or disprove that $f^{-1} \in \mathcal{R} [c,d]$.

Remark: I tried to use integration by parts to find $\int_{c}^{d} f^{-1}$ and to prove that was the right limit of Riemann sum, but failed. But I think this idea might be useful.

Solution 1:

Let $C\subset[0,1]$ be the middle thirds Cantor set, and let $D\subset[0,1]$ be a fat Cantor set. Define $f:[0,1]\to[0,1]$ such that $f\vert_C$ is an order preserving homeomorphism of $C$ onto $D$ (by mapping to corresponding endpoints of the removed intervals), and $f\vert_{[0,1]\setminus C}$ is an order reversing homeomorphism of $[0,1]\setminus C$ onto $[0,1]\setminus D$ (by mapping linearly to corresponding removed intervals and then composing with $x\mapsto 1-x$). Then $f$ is discontinuous at each point of $C$ and continuous at each point of $[0,1]\setminus C$. Since $C$ has measure zero, $f$ is Riemann integrable. On the other hand, $f^{-1}$ is discontinuous at each point of $D$, so it is not Riemann integrable.

If you don't want to appeal to the Lebesgue criterion of Riemann integrability, you could work explicitly with Riemann sums of $f$ and $f^{-1}$. In the case of $f$, you can choose partitions such that the contribution of intervals containing points of $C$ is arbitrarily small. In the case of $f^{-1}$, the values of $f^{-1}$ on $D\cap[\frac{1}{2},1]$ are at least $\frac{2}{3}$ and the values of $f^{-1}$ on $[\frac{1}{2},1]\setminus D$ are at most $\frac{1}{2}$. Every interval contains points from $[0,1]\setminus D$, so the difference between upper and lower sums will always be at least $(\frac{2}{3}-\frac{1}{2})(\frac{1}{2}m(D))\gt 0$.