Suppose that $F$ is a distribution function that is absolutely continuous with respect to Lebesgue measure on $\mathbb{R}$ with density $f$. Let $F^{-1}$ be the associated quantile function and assume that $F^{-1}(t)$ is a singleton for all $t \in (0,1)$ (e.g. as in a normal distribution).
I am interested in properties of the derivative of $F^{-1}$, call it $G$. From calculus we expect that $G(t) = 1/f(F^{-1}(t))$ as long as $f(F^{-1}(t)) \neq 0$.
However, $f$ is only unique a.e. with respect to Lebesgue measure. So supposedly I could choose $f$ at $F^{-1}(t)$ to make $G(t)$ any value including non-existent by choosing $f(F^{-1}(t)) = 0$. This seems absurd. Can you explain to my why it is not absurd and/or clear up my confusion regarding measure-theoretic probability that is leading me to this conclusion?
We don't "expect" this; we know (from the inverse function theorem) that the derivative of $F^{-1}$ is equal to $1/F'(F^{-1}(t))$ as long as $F'(F^{−1}(t))\ne 0$.
This is what the calculus theorem says, and there is no $f$ in the above statement. Sure, if you denote the derivative $F'$ by $f$, you can write the above formula as $1/f(F^{-1}(t))$. But if $f$ is some other function, you cannot do that.
When people say "redefining a function on a set of measure zero does not change anything", they mean for the purpose of integration. Differentiation is another thing.