Application of Fubini's theorem without integrals

Question

Suppose $$ \forall s \in [0,t] : \mathbb{P}\Big( \omega \in \Omega : \text{a property depending on $s$ and $\omega$ does hold}\Big)=1. $$ Why does Fubini's theorem imply that $$ \big(\lambda \vert_{[0,t]} \otimes \mathbb{P}\big)\Big( (s,\omega) \in [0,t] \times \Omega : \text{the same property depending on $s$ and $\omega$ does hold}\Big)=t $$ ?

Answer 1

Let the function $$ F\colon \left(s,\omega\right)\in [0,t]\times \Omega\mapsto \begin{cases} 1&\mbox{ if the property depending on }s \mbox{ and }\omega\mbox{ holds,}\\ 0&\mbox{ otherwise}. \end{cases} $$ Provided that this function is measurable, one uses Fubini's theorem to get $$ \int_{[0,t]\times \Omega}F\left(s,\omega\right)\mathrm d\lambda\left(s\right) \otimes \mathbb P\left(\omega\right) =\int_{[0,t]}\left(\int_\Omega F\left(s,\omega\right)\mathrm d\lambda\left(s\right)\right) \mathbb P\left(\omega\right)$$ and concludes noticing that the inner integral is equal to $1$.