Same occupation measure $\Rightarrow$ same trajectory

Question

Let $f$ be a $\mathcal{C}^1$ vector field (VF) on a compact subset $M \subset \mathbb{R}^n$. $M$ inherits the Euclidean metric. We define a dynamical system by $$\dot{x}(t)=f(x(t))$$ The occupation measure is defined as $$ \mu_{x_{0}, T}^f(B) \, = \, \frac{1}{T} \int_{0}^{T} \mathbb{1}_{B}(x(s)) \, d s $$

where $x(t)$ is a trajectory with $t \in [0, T]$, starting from $x_0$, $B \subset \mathbb{R}^n$ is some Borel measurable set, and $\mathbb{1} \,$ denotes an indicator function. By $\Gamma_{x_0}^f=\{x(t)|t \in T\}$ we denote a trajectory.

If two vector fields $f,g$ have equal occupation measures for all times $T$, can we conclude that the underlying trajectories are equal? In other words: $$ \mu_{x_{0}, T}^f(B) =\mu_{x_{0}, T}^g(B) \ \forall T \in \mathbb{R}, \forall B \quad \Rightarrow \quad \Gamma_{x_0}^f=\Gamma_{x_0}^g $$

Answer 1

Under your hypotheses, you'll have (for each bounded continuous function $\varphi$) $$ \int_0^T \varphi(x^f(s))\,ds = \int_0^T \varphi(x^g(s))\,ds,\qquad\forall T\ge 0, $$ where $x^f$ and $x^g$ are the trajectories corresponding to $f$ and $g$. Differentiating with respect to $T$: $$ \varphi(x^f(T)) = \varphi(x^g(T)),\qquad\forall T\ge 0. $$ As the bounded continuous functions separate points, you must have $$ x^f(T) = x^g(T),\qquad\forall T\ge 0. $$