Height of saltus of Poisson process at points of discontinuity In my lecture notes a Poisson process is defined as follows:

Suppose that $\vartheta: [0,\infty) \to [0,\infty)$ is a positive Borel-measurable function satisfying $\int_0^T \vartheta(t) dt < \infty$ for all $T > 0$. Then an $\mathbb N_0$-valued process $N = \{ N_t \}_{t\in [0,\infty)}$ is said to be a Poisson process with intensity $\vartheta$ if
(P1) $N$ is a counting process, i.e. right-continuous and increasing
(P2) $N(0) = 0$ a.s.
(P3) $N(t) - N(s)$ is independent of $\mathfrak F^N(s)$ for all $s \le t$,
(P4) $N(t) - N(s) \sim Poisson(\int_s^t \vartheta(u) du)$ for $s \le t$.

Does this force the saltus $N(t) - \sup_{s < t} N(s)$ at the points of discontinuity to be exactly $1$? For many constructions this seems to be assumed, for example in an exercise the distribution of 
$$
 \tau := \inf\{ t \ge 0 : N_t = 1 \}
$$
should be determined, but this is just defined if the saltus is exactly one, for otherwise if it would jump two units at once in its first jump then $\tau$ would not be defined. Also that there is some space $t > s$ if $N_s - N_t > 0$ is used when I write $\{ \tau \le s \} = \{ N_s \ge 1 \}$ by also using that it is increasing and right-continuous, surely if this does not hold then it could be in some way fixed by setting
$$
 \tau := \inf\{ t \ge 0 : N_t \ge 1 \}
$$
but that is not the way it is defined in the first place. Otherwise in all definitions I do not found a reference to the saltus at the points of discountinuity. So what could be said about this saltus, and how to resolve the above technicalities?
 A: First note that if $Y$ has the Poisson distribution with parameter $\lambda$ then
$P[Y\ge 2]=(1/2)\lambda^2+o(\lambda^2)$ as $\lambda\to 0$. In particular, there exists $\lambda_0>0$ such that $P[Y\ge 2]\le \lambda^2$ for $0<\lambda<\lambda_0$.
Write $\Lambda(t):=\int_0^t\vartheta(s)\,ds$.
Let's show that on the time interval $[0,1]$, $N$ can only have jumps of size $1$, almost surely.In fact
$$
\{N|_{[0,1]} \hbox{ has a jump of size 2 or more}\} =\cap_{m=1}^\infty\cup_{k=1}^m\{N(k/m)-N((k-1)/m)\ge 2\}.
$$
The function $\Lambda$ defined above is continuous, hence uniformly continuous on $[0,1]$. Thus, given $\epsilon\in(0,\lambda_0)$, there exists an integer $m_0$ so large that if $s,t\in[0,1]$ with $|t-s|<1/m_0$ then $|\Lambda(t)-\Lambda(s)|<\epsilon$.
Thus, if $m\ge m_0$ then
$$
P(N(k/m)-N((k-1)/m)\ge 2)\le[\Lambda(k/m)-\Lambda((k-1)/m)]^2\le\epsilon\cdot [\Lambda(k/m)-\Lambda((k-1)/m)],
$$
and so
$$
\eqalign{
P(\cup_{k=1}^m\{N(k/m)-N((k-1)/m)\ge 2)
&\le\sum_{k=1}^mP(N(k/m)-N((k-1)/m)\ge 2)\cr
&\le\sum_{k=1}^m[[\Lambda(k/m)-\Lambda((k-1)/m)]^2\cr
&\le\epsilon\cdot\sum_{k=1}^m[\Lambda(k/m)-\Lambda((k-1)/m)]\cr
&\le \epsilon\cdot\Lambda(1).\cr
}
$$
It follows that 
$$
P(\{N|_{[0,1]} \hbox{ has a jump of size 2 or more}\})=0.
$$
