Prove the critical points of $\psi$ form a zero set (counterexample?)

In Pugh's Real Mathematical Analysis (first edition), ch. 3, exercise 41 (b) (p. 195):

Assume that $\psi: [c, d] \to [a, b]$ is continuously differentiable. Prove that if $f \circ \psi$ is Riemann integrable for each Riemann integrable $f$ on $[a, b]$, then the critical points of $\psi$ form a zero set.

But if $\psi$ is constant, then it is everywhere critical and thus its critical points are certainly not a zero set (assuming $c < d$ of course). Yet $f \circ \psi$ is then constant for all $f$, and thus Riemann integrable. So this seems to be a counterexample. Have I done something wrong?

• Your reasoning seems just fine to me. There are no other assumptions on $\psi$? Perhaps $\psi$ is a bijection between these two intervals?
– zhw.
May 21, 2016 at 3:38
• Nope, no other assumptions. Certainly if $\psi$ is a bijection this seems to work, since one could construct a function $f$ discontinuous on $\psi$'s critical values, which would form a zero set (as they do for all cont. differentiable functions), and then $f$ would be integrable but $f \circ \psi$ would be discontinuous at $\psi$'s non-zero set of critical points. Actually, that's how I tried to prove this at first, but I couldn't prove $f \circ \psi$ was actually discontinuous at those points! I'll bet Pugh meant to include that assumption. Thanks for the sanity check. May 21, 2016 at 3:50