Assume that $f:[a,b]\to [c,d]$ is a Riemann integrable bijective function.
Is $f^{-1}$ also Riemann integrable?
I think it's true. I have tried to find a counterexample but failed since bijective is so strong.
But I don't know where should I started with.
The answer is no. Here is an example.
Let $K$ and $L$ be to closed nowhere dense subsets of $[0,1]$ without isolated points, both containing $0$ and $1$, such that $K$ has Lebesgue measure $0$ and $L$ has non-zero Lebesgue measure.
Let $\psi : L\to K$ be any bijection which is discontinuous at every point. (Such things should exist because $L$ has no isolated points; I don't have an obvious example right now).
The sets $U:=[0,1]\setminus K$ and $V:=[0,1]\setminus L$ are open subsets of $\mathbb R$; write them as countable unions of pairwise disjoint open intervals, say $U=\bigcup_{n\geq 0} I_n$ and $V=\bigcup_{n\geq 0} V_n$. For each $n\geq 0$, choose an affine bijection $\phi_n :I_n\to J_n$.
Now define a bijection $f:[0,1]\to [0,1]$ as you guess: $f\equiv \phi_n$ on each $I_n$ and $f\equiv \psi^{-1}$ on $K$. Then, $f$ is continuous at each point $x\in [0,1]\setminus$, so it is Riemann-integrable on $[0,1]$ because $K$ has measure $0$. On the other hand, $f^{-1}$ is discontinuous at every point $y\in L$ because its restriction to $L$ (namely $\psi$) is already discontinuous at every point; hence $f^{-1}$ is not Riemann-integrable because $L$ has non-zero Lebesgue measure.
Edit I discovered the link given by Sarastro just after writing this answer!
