Evaluate $P(\bigcap_{t>0}\bigcup_{i=0}^1 \{\lim_{n\to\infty}(N(t)-N(t-\frac{1}{n}))=i\})$ where $N(t)$ is a Poisson process $N(t)$ is a poisson process with constant parameter $\lambda$, please evaluate
$$P(\bigcap_{t>0}\bigcup_{i=0}^1 \{\lim_{n\to\infty}(N(t)-N(t-\frac{1}{n}))=i\})$$
It's easy to derive $P(\{\lim_{n\to\infty}(N(t)-N(t-\frac{1}{n}))=0\})=1$, but I'm confused   with the joint. I think it should be $1$ since in most cases the process will be at most  increased by $1$.
Edit: the original question put it in the form
$$P(\{\bigcup_{i=0}^1 \{\lim_{n\to\infty}(N(t)-N(t-\frac{1}{n}))=i\},\forall t>0)$$
I don't know whether there is a difference.
 A: The probability is indeed $1$. Note that $\bigcap_{t>0}\bigcup_{i=0}^1\{\lim_{n\to\infty}(N(t)-N(t-1/n))=i\}$ is the event that all jumps are of size $1$. To prove that this event has probability $1$ we will show that the complement has probability $0$. We will use the standard notation $\Delta N(t):=\lim_{n\to\infty}(N(t)-N(t-1/n))$ to denote the jump process at time $t>0$.
Now, if $t>0$ such that $\Delta N(t)\geq2$ we may take $n\in\mathbb{N}$ such that $n\geq t$. For any $k\in\mathbb{N}$ we can find $i\in\{1,\dotsc,n2^k\}$ such that $t\in((i-1)/2^k,i/2^k]$. We note that $N(i/2^k)-N((i-1)/2^k)\geq\Delta N(t)\geq2$. So we may write
$$
\{\exists t>0\text{ such that }\Delta N(t)\geq2\}=\bigcup_{n=1}^\infty\bigcap_{k=1}^\infty\bigcup_{i=1}^{n2^k}\{N(i/2^k)-N((i-1)/2^k)\geq 2\}.
$$
To show that the right-hand side has probability $0$ it is sufficient to show that
$$
P(\bigcap_{k=1}^\infty\bigcup_{i=1}^{n2^k}\{N(i/2^k)-N((i-1)/2^k)\geq 2\})=0
$$
for each $n\in\mathbb{N}$. Moreover, since
$$
P(\bigcap_{k=1}^\infty\bigcup_{i=1}^{n2^k}\{N(i/2^k)-N((i-1)/2^k)\geq 2\})\leq P(\bigcup_{i=1}^{n2^{k_0}}\{N(i/2^{k_0})-N((i-1)/2^{k_0})\geq 2\})
$$
for each $k_0\in\mathbb{N}$ it suffices to show that the right-hand side converges to $0$ as $k_0\to\infty$.
To this end we see that
$$
\begin{align*}
P(\bigcup_{i=1}^{n2^{k_0}}\{N(i/2^{k_0})-N((i-1)/2^{k_0})\geq 2\})&\leq\sum_{i=1}^{n2^{k_0}}P(N(i/2^{k_0})-N((i-1)/2^{k_0}\geq 2) \\
&=n2^{k_0}P(N(1/2^{k_0})\geq2).
\end{align*}
$$
Finally we use fact that $N(1/2^{k_0})\sim\mathrm{Poi}(\lambda/2^{k_0})$. We find that
$$
n2^{k_0}P(N(1/2^{k_0})\geq 2)=n2^{k_0}(1-(e^{-\lambda/2^{k_0}}+\lambda/2^{k_0}e^{-\lambda/2^{k_0}}))\to0\quad\text{as }k_0\to\infty.
$$
This finishes the proof.
