definition $q_-(\zz)=1$ if $\zz \in A$, $q_-(\zz)=0$ if $\zz \in
B$, and $0\le q_-(\zz)\le1$ otherwise. The backward committor function is useful because by the Markovian assumption it follows that, at statistical steady state, the probability density to observe a trajectory at point $\zz \in \Omega$ at any given time $t$ *and* that this trajectory is assigned to $A$ at that time is simply $\varrho_{A}(\zz)= \varrho(\zz) \, q_-(\zz)$. Similarly the probability density to observe a trajectory at point $\zz \in \Omega$ at time $t$ *and* that this trajectory is assigned to $B$ at that time is $\varrho_{B}(\zz) = 1-\varrho_A(\zz)= \varrho(\zz) \,
(1-q_-(\zz))$. Since $\rho_A = \int_\Omega \varrho_{A}(\zz)d\zz$ and $\rho_B = \int_\Omega \varrho_B(\zz)d\zz$ by definition, this implies that $$\label{eq:rhoAB}
  \rho_A=\int_{\Omega} \varrho(\zz) \, q_-(\zz) \, d\zz \le 1, \qquad
  \rho_B=1-\rho_A.$$ Similarly, it is easy to see that $\nu_R$ is the total probability flux associated with $\varrho_{A}(\zz)= \varrho(\zz) \, q_-(\zz)$ going through any dividing surface between $A$ and $B$ (like e.g. the boundary of $B$). The explicit form of this flux depends on the specifics of the dynamics and it is given in Ref. : let us omit to repeat this formula here since it will not be important in what follows.

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