I am trying to prove a corollary to the Riemann Lebesgue Theorem in Pugh's Real Mathematical Analysis. For context, the Riemann Lebesgue Theorem states that:
"A function $f:[a,b]\to\mathbb{R}$ is Riemann integrable iff it is bounded and its set of discontinuity points is a zero set."
The corollary I am trying to prove on page 182 states that:
"If f is Riemann integrable and $\psi$ is a homeomorphism whose inverse satisfies a Lipschitz condition, then $f\circ\psi$ is Riemann integrable."
Let $f:[a,b]\to\mathbb{R}$ and $\psi:[c,d]\to[a,b]$. Then to prove this corollary, we just need to show that $f\circ\psi$ is bounded and its set of discontinuity points is a zero set. Boundedness is trivial and my main question is about showing why the set of discontinuity points, $D_{f\circ\psi}$ is a zero set. Since $f$ is Riemann integrable, its set of discontinuity points $D_f$ is a zero set. My intuition is that since $\psi$ is a homeomorphism, $\psi^{-1}$ is a bijection that puts the set of discontinuity points $D_f$ to a one-to-one correspondence with $D_{f\circ\psi}$. This would allow us to conclude that $D_{f\circ\psi}$ is also a zero set. However, my intuition doesn't use the fact that $\psi^{-1}$ satisfies a Lipschitz condition, which makes me suspicious that I am making incorrect assumptions in my intuition. Could you help me pinpoint why we need $\psi^{-1}$ to satisfy a Lipschitz condition?
If $\psi^{-1}$ is not Lipschitz, there could be a closed set $E$ of measure zero with $\psi^{-1}(E)$ of positive measure.
For example, use a strictly increasing homeomorphism to map the middle-thirds Cantor set onto a fat Cantor set.
Take $f$ to be the indicator function $E$ (so $f$ is integrable). Then $f\circ\psi$ is the indicator function of $\psi^{-1}(E)$ so it is not integrable.