If \mu\equiv 0 then N\equiv 0 satisfies the conditions i)–iii). Otherwise, in the case of
finite measure \mu, given Z, a
Poisson random variable with rate \mu(E), and X_{1}, X_{2},\ldots, mutually
independent random variables with
distribution \frac{\mu}{\mu(E)}, define N_{\cdot}(\omega) = \sum\limits_{i=1}^{Z(\omega)} \delta_{X_i(\omega)}(\cdot) where \delta_{c}(A) is a
degenerate measure located in c. Then N will be a Poisson random measure. In the case \mu is not finite the
measure N can be obtained from the measures constructed above on parts of E where \mu is finite. ==Applications==