Let $\xi$ be a uniformly random variable on $[0,1]$ defined on some probability space $(\Omega,\mathcal{F})$. Define the process $\xi_t:=\min(\xi,t)$ for $0\le t\le 1$. And let $\mathcal{F}_t=\sigma(\xi_s, 0\le s\le t)$, it is easy to see that $\mathcal{F}_0$ satisfies the $0$-$1$ law. Now let
$$\mathcal{F}_t^+=\cap_{s>t}\mathcal{F}_s$$
My question is whether $\mathcal{F}_0^+$ satisfies $0$-$1$ law? Thx a lot!
There is no need to talk about "some probability space" on which your random variable is defined (as probabilists like to do). Everything is determined just by the uniform (aka Lebesgue) measure on $[0,1]$. For this space $F_t$ is just the $\sigma$-algebra generated by measurable subsets of $[0,t]$. The rest is an obvious exercise.
The random variables $\xi_t$ have a big atom, since $P(\xi_t=t)=P(\xi\ge t)=1-t$. So the probabilities of all events in $\sigma(\xi_t)$ are contained in the intervals $[0,t]$ and $[1-t,1]$. Since an event of $\mathcal{F}_0^+$ has to be in $\sigma(\xi_t)$ for all $1\ge t>0$, it follows that its probability can only be $0$ or $1$.
Yes, because since $ P(\xi=0)=0$, immediately after time 0 the process is just the deterministic identity function $\xi_t=t $.
