In the following, all the measure spaces are endowed with the Borel $\sigma$-algebra corresponding to their topology (we take the usual topology on $[0,\infty)$).
Let $E$ be a Polish space and let $\nu$ be a $\sigma$-finite measure on $E$ .
Let $\psi$ be a Poisson random measure on the space $E \times [0,\infty)$ with intensity measure $\nu \times dx$ (where $dx$ denotes the Lebesgue measure).
It is often said that $$\psi ( E \times \{t\}) \in \{0,1\} \hspace{30pt} (1)$$ almost surely, for all $t \geq 0$. (This allows us to think of the random measure as a dynamic process taking values in $E \cup \{\gamma\}$ where $\gamma$ is an isolated point).
But I don't understand this, because
$$ (\nu \times dx ) ( E \times \{t\}) = \nu(E) \cdot dx(\{t\}) \leq \infty \cdot 0 = 0 $$
so that the number of arrivals at time $t$ has a Poisson distribution with rate $0$. That is, $0$ points arrive with probability $1$.
What is the problem with this reasoning, and how is $(1)$ proved correctly?
Many thanks for your help in advance.