If $\tau$ is a stopping time with $\text P[\tau>s+t]=\text P[\tau>s]\text P[\tau>t]$, how do we determine the rate of the exponential distribution?

Solution 1:

$A \to \pi_{\omega}(A \times B)$ is the random measure corresponding to a Poisson process on $[0,\infty)$. The rate of this Poisson process, is also the rate of the exponential waiting time for the first point of the process. See Jumps of Lévy process. So what is being used here is a simple but powerful property of the Poisson process on a halfline: The rate $\lambda$ of the exponential waiting time for the first point of the process, equals the expected number of points of the process that land in the unit interval. https://en.wikipedia.org/wiki/Poisson_point_process#Interpreted_as_a_point_process_on_the_real_line