The following proof, that how to derive Poisson Distribution from a Poisson Process, is from my textbook, Elementry Probability Theory(Fourth Edition), written by Kai Lai Chung, Farid AitSahlia.
Proof
Define $T_j$ as interarrival time: $$P(T_j > t) = e^{- \alpha t}, t \ge 0$$ $\alpha$ is intensity of the flow.
Define $S_n$ as the waiting time until the nth arrival: $$S_n = \sum_{j=1}^n T_j$$
Define $N(t)$ as the total number of arrivals in the time interval $[0, t]$. By definition, we have $$\{ N(t) \ge n \} = \{ S_n \le t \}$$
Taking the difference of both sides in the above equations for $n$ and $n+1$, we obtain $$\{ N(t) = n \} = \{ S_n \le t \} - \{ S_{n+1} \le t \} = \{ S_n \le t < S_{n+1} \} (1)$$
Then, $$P\{ N(t) = n \} = P\{ S_n \le t \} - P\{ S_{n+1} \le t \} (2)$$
(Here omitted)
Finally, we get $P \{ S_n \le t\} = \int_0^t \frac{\alpha^n}{(n-1)!} u^{n-1} e^{- \alpha u}$(Erlang distribution), $P\{ N(t) = n \} = \frac{\alpha^n}{n!} t^n e^{-\alpha t}$(Poisson distribution).
Question
Later on is the rest of a complete proof of Poisson process to get Poisson Distribution. But here is my question! How to get equation $(2)$ from the equation $(1)$?
Recalling the definition of probability theory, event needs to exist as a subset of a sample space $\Omega$, and $P( \Omega ) = 1$. Only events in a same sample space can be operated. Looking at the probabilities in equation $(2)$, I think the events are in different sample space? $\{ N(t) = n \}$ in $\{ \text{variable n under certain t} \}$, $\{ S_n \le t \}$ in $\{ \text{variable t under certain n} \}$, $\{ S_n \le t \}$ in $\{ \text{variable t under certain n+1} \}$. I choose them by an important condition that $P(\Omega) = 1$.
Why could I operate events in different sample spaces?
They are in one sample space. The sample space is all the possible records. Trying to be more precise, each record can be represented by a sequence of nonnegative real numbers: $t_1, t_2, t_3, \cdots$
Thanks to @drhab and @Did to comment, they unraveled my confusion. But I've still writen this answer, because I hope every question has an answer.