Let $E$ be a normed $\mathbb R$-vector space, $(X(t))_{t\ge0}$ be an $E$-valued càdlàg Lévy process on a probability space $(\Omega,\mathcal A,\operatorname P)$ and$^1$ $$N(t,A):=\left|\left\{s\in(0,t]:\Delta X(s)\in A\right\}\right|\;\;\;\text{for }t\ge0$$ for some $A\in\mathcal B(E)$ with $0\not\in\overline A$.
I want to show that $(N(t,A))_{t\ge0}$ has independent increments. A proof of this claim can be found in Theorem 2.3.5-(2) in the book of Applebaum, but I don't get his argumentation:
Regarding (2.3): Since $0\not\in\overline A$, $$r:=\operatorname{dist}(0,A)>0$$ and hence $$\left|\left\{t\in[a,b]:\Delta x(t)\in A\right\}\right|\le\left|\left\{t\in[a,b]:\left\|\Delta x(t)\right\|_E\ge r\right\}\right|\tag1$$ is finite for all $a,b\in\mathbb R$ and every càdlàg $x:[0,\infty)\to E$. But this only yields the existence of the $t_i$ (as claimed in the proof) which depend on $\omega$ or am I missing something?
Regarding (2.4): I actually don't get this. If $x:[0,\infty)\to E$ has a left-limit $t\ge0$, then it clearly holds that for every $\varepsilon>0$, there is a $\delta>0$ with $$\left\|x(r)-x(t)+\Delta x(t)\right\|_E=\left\|x(r)-x(t-)\right\|_E<\varepsilon\tag2$$ for all $t\in(t-\delta,t)$. But this doesn't yield the "only if" part of the claim in the proof, since we cannot choose $a=\Delta X(\omega,t)$ (for some fixed $\omega)$, since we only know (by assumption) that $\Delta X(\omega,t)\in A$, which doesn't necessarily imply that $-a\in A$.
And the "if" part isn't clear to me at all.
How can we show these things rigorously?
EDIT: Assume $(X(t))_{t\ge0}$ is a Lévy process with respect to a filtration $(\mathcal F_t)_{t\ge0}$. Let $\tau_0:=0$. Then $$\tau_n:=\inf\{t>\tau_{n-1}:\Delta X(t)\in A\}$$ is a stopping time with respect to the right-continuous filtration $(\mathcal F_t^+)_{t\ge0}$. Moreover, $$N(t,A)=\sum_{n\in\mathbb N}1_{[0,\:t]}(\tau_n)\tag3$$ for all $t\ge0$. Now $$N(t,A)-N(s,A)=\sum_{n\in\mathbb N}1_{(s,\:t]}(\tau_n)\tag4$$ for all $t\ge s\ge0$. Since $1_{(s,\:t]}(\tau_n)$ is $\mathcal F^+_t$-measurable, for all $t\ge s\ge0$, $N(t,A)-N(s,A)$ is $\mathcal F^+_t$-measurable as well. But does this help?
As usual, $x(t-):=\lim_{s\to t-}x(s)$ and $\Delta x(t):=x(t)-x(t-)$ for every $x:[0,\infty)\to E$ with a left-limit at $t\ge0$.
For (2.3), yes it depends on $\omega$. Also I think it should be "$N(t,A)-N(s,A)\geq n$ if and only if there exists..." which makes more sense to me, and later we show instead $\bigl(N(t,A)-N(s,A)\geq n\bigr)\in\sigma\bigl(X(v)-X(u):s\leq u<v\leq t\bigr)$. This is still enough for proving independence as $$\bigl(N(t,A)-N(s,A)=n\bigr)=\bigl(N(t,A)-N(s,A)\geq n\bigr)\cap\bigl(N(t,A)-N(s,A)\geq n+1\bigr)^{c}.$$
For (2.4), there's a typo in it, it should be "$|X(u)-X(w)-a|<\varepsilon$". See the errata here. The statement "for every $\varepsilon>0$, there is a $\delta>0$ such that $0<u-w<\delta$ $\Rightarrow$ $|X(u)-X(w)-a|<\varepsilon$" is the exact definition of $\lim_{t\uparrow\uparrow u}X(t)=X(u)-a$. Also note that $\Delta X(u)=a$ $\iff$ $X(u)-\lim_{t\uparrow\uparrow u}X(t)=a$, hence the if and only if statement holds.