Jump size of process in proof of martingale CLT bounded

I'm currently trying to understand the proof of the final theorem of (Helland, I. S. (1982). Central limit theorems for martingales with discrete or continuous time) where I need the following: For $$\{M(t)\}_{t\geq0}$$ a càdlàg local martingale we define $$a^\epsilon[M](t)=\sum_{s\leq t}|\Delta M(s)| 1\{|\Delta M(s)|>\epsilon\}$$ and denote it's compensator by $$â^\epsilon[M]$$.

We then define $$\overline{M}^\epsilon:=a^\epsilon[M]-â^\epsilon[M]$$ and $$\underline{M}^\epsilon:=M-\overline{M}^\epsilon$$.

"It can be shown that the maximal jump in $$\underline{M}^\epsilon$$ has size less or equal than $$2\epsilon$$."

Why is that? I also tried looking up in the references but ended up with "LE THEOREME FONDAMENTAL SUR LES MARTINGALES LOCALES" by Meyer which says something about "totally inaccessible times", on which I couldn't find enough to make sense of the proof (which is written in French, which I don't speak…)

The process $$b^\epsilon[M](t): = M (t) - a^\epsilon[M](t)$$ does not have jumps larger than $$\epsilon$$. We have
$$\underline{M}^\epsilon (t) = M-\overline{M}^\epsilon (t) = M (t) - a^\epsilon[M] (t) + â^\epsilon[M] (t)$$ $$= b^\epsilon[M](t) + â^\epsilon[M] (t).$$
So we need to show that $$â^\epsilon[M]$$ does not have jumps of size $$> \epsilon$$. Assume that it does; then either there exists a stopping time (with respect to $$\{\mathscr{F}(t)\}]$$) $$\tau$$ such that $$P\{\Delta â^\epsilon[M] (\tau) > \epsilon\} >0, \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$ or there exists a stopping time $$\tau$$ such that $$P\{\Delta â^\epsilon[M] (\tau) < - \epsilon\} >0. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$$ Assume (1), and let $$Q = \{\Delta â^\epsilon[M] (\tau) > \epsilon\}$$, $$P(Q) >0$$. Then $$Q$$ is $$\mathscr{F}(\tau-)$$-measurable since $$â^\epsilon[M]$$ is predictable, $$E [\Delta a^\epsilon[M] (\tau)|Q] > \epsilon$$, and $$E [\Delta M (\tau)|Q] = E [\Delta a^\epsilon[M] (\tau)|Q] + E [\Delta b^\epsilon[M] (\tau)|Q] > \epsilon - \epsilon = 0.$$ This is a contradiction since $$\tau$$ is predictable, and so $$E [\Delta M (\tau)|Q] = 0$$.