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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…)

Thanks a lot in advance.

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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 $E [\Delta M (\tau)|Q] = 0$.

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