A measurable set that acts as a speedometer Definitions and some motivation:
Say a car is driving on a straight road. All we know is where it starts, and how much time it spends in certain stretches of the road. With just this much information, can we pinpoint the exact trajectory of the car?
Translating this loosely to math - does there exist a measurable subset $S$ of $\mathbb R$ such that any absolutely continuous $f: [0, \infty) \to \mathbb R$ with $f(0) = 0$ is determined entirely by how much time it spends in $S$?
More precisely, denote by $\mathcal C_0$ the subset of absolutely continuous, real valued functions $f$ on $[0, \infty)$ such that $f(0) = 0$.
For any measurable subset $S$ of $\mathbb R$, and
any $f \in \mathcal C_0$, define the function $T(f): [0, \infty) \to [0, \infty)$ by
$T(f)(x) := \mu(\{t|\ t \in [0, x), f(t) \in S \})$.
Question:
Does there exist a measurable subset $S$ of $\mathbb R$ such that the operator on $\mathcal C_0$ sending $f$ to $T(f)$ is injective?
Note: Here $\mu$ denotes the usual Lebesgue measure.
 A: No, $T$ cannot be injective.

If $S \cap (a, b)$ has zero Lebesgue measure, then $T$ will assign the same value to $f_1$ and $f_2$ if, say, $f_1(t) = f_2(t)$ for $t < 1$ and $f_1(t), f_2(t) \in (a, b)$ for $t \geqslant 1$, and $f_1'(t) \ne 0$ and $f_2'(t) \ne 0$ for $t \geqslant 1$.
Suppose that $S \cap (a, b)$ has a positive Lebesgue measure for every interval $(a, b)$, and write $$\phi(x) = \int_0^x \mathbb 1_S(y) dy$$ (that is, $\phi(x) = \mu(S \cap [0, x))$ for $x \geqslant 0$ and $\phi(x) = -\mu(S \cap [x, 0))$ for $x < 0$). Then $\phi$ is strictly increasing, and therefore the inverse function $\psi = \phi^{-1}$ is well-defined. We claim that $\psi(t) \in S$ for almost all $t \in \mathbb R$. This shows that if $f_1(t) = \psi(t)$ and $f_2(t) = \psi(-t)$, then $T(f_1) = T(f_2)$. Thus, it remains to prove our claim.

Observe that
$$ t = \phi(\psi(t)) = \int_0^{\psi(t)} \mathbb 1_S(y) dy = \int_0^t \mathbb 1_S(\psi(s)) d\psi(s) . $$
Therefore, $\mathbb 1_S(\psi(s)) \ne 0$ for almost all $s$ with respect to $d\psi$. Furthermore, since $0 < \phi(y) - \phi(x) \leqslant y - x$ when $x < y$, we have $\psi(t) - \psi(s) \geqslant t - s$ when $s < t$. In particular, the Lebesgue measure is absolutely continuous with respect to $d\psi$. It follows that $\mathbb 1_S(\psi(s)) \ne 0$ for almost all $s$ also with respect to the Lebesgue measure, as claimed.
