Given a dynamical system, the most common definition of a physical measure is, that the measure $\mu$ is invariant w.r.t. to the flow $\Phi(t,x)$ and its basin
$$ B(\mu)=\left\{x \in M | \lim _{t \rightarrow \infty} \frac{1}{t} \int_{0}^{t} \psi\left(\Phi(s,x)\right) d s=\int \psi d \mu,\quad \forall \psi \in C^{0}(M)\right\}$$
has a postivie Lebesgue measure on $M$ (which we assume to be a smooth, compact manifold).
However, I also read the following definition. The occupation measure is defined by $$ \mu_{T, x_{0}}(S)=\frac{1}{T}\mu \left(\left\{t \in \left[0, T\right] \mid \Phi\left(t, x_{0}\right) \in S\right\}\right) =\frac{1}{T} \int_{0}^{T} \mathbb{I}_{B}(x(s)) d s $$ where $S$ is some Borel measurable set. Then, a measure $\mu$ is physical if there is a set $B$ with positive Lebesgue measure, such that for Lebesgue almost every $x \in B$, as $T \rightarrow \infty$, the occupation measure $\mu_{x, T}$ converges weakly to $\mu$.
My question is the following: Are these two definitions equivalent or in which way are they related? Which role does the occupation measure play?
The occupation measure $\mu_{x,T}$ converging weakly to $\mu$ means that for all continuous functions $\psi \in C^0(M)$, $$ \int \psi d\mu_{x,T} \underset{T \rightarrow \infty}{\longrightarrow} \int \psi d\mu $$ but $$ \int \psi d\mu_{x,T} = \frac{1}{T} \int_0^T \psi(x(s))ds $$ so the definitions are exactly the same, where $B$ from the second definition is taken to be contained in $B(\mu)$