Let
- $E$ be a normed $\mathbb R$-vector space
- $x:[0,\infty)\to E$ be càdlàg
- $x(t-):=\lim_{s\to t-}x(s)$ and $\Delta x(t):=x(t)-x(t-)$ for $t\ge0$
- $\tau^B_0:=0$, $I^B_n:=\{t>\tau^B_{n-1}:\Delta x(t)\in B\}$ and $\tau^B_n:=\inf I^B_n$ for $n\in\mathbb N$ and $B\subseteq E$
- $t\ge0$
- $B\subseteq E$ with $0\not\in\overline B$
- $r:=\operatorname{dist}(0,\overline B)$
- $I:=\{s\in(0,t]:\|\Delta x(s)\|_E\ge r\}$
Since$^1$ $x$ is càdlàg, $$I=\{\tau_1,\ldots,\tau_n\}\tag1$$ for some $n\in\mathbb N_0$ and $\tau_1,\ldots,\tau_n\in(0,t]$ with $\tau_1<\cdots<\tau_n$.
Question: How are $\tau_1,\ldots,\tau_n$ related to $(\tau^B_i)_{i\in\mathbb N}$? Can we show $\tau_i=\tau^B_i$ for all $i\in\{1,\ldots,n\}$? Maybe at least when $B$ is closed?
It clearly holds $B\subseteq B_r(0)^c$. So it should hold $\tau_i\ge\tau^B_i$ for all $i\in\{1,\ldots,n\}$.
$^1$ A regular function on a bounded interval has finitely many jumps exceeding a fixed size.