The following is a question about a notation that Protter uses in the proof of the Ito's formula for cadlag processes of finite variation (FV) that appears on Stochastic Integration and Differential Equations at page 80.

The Ito's formula is this enter image description here

And my doubt is that I have not understood yet what does the sets $A = A(\epsilon, t)$ and $B = B(\epsilon, t)$ mean in order to conclude that $A \cup B $ equals the set of stopping times of $X$ on $(0,t]$. The definition of these two sets is the following

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For me the definition of $ A= A(\epsilon, t)$ (as a set that depends on $\epsilon $ and $t$) should be: let $A(w) = A(\epsilon, t)(w)$ be any finite set of jump times of $X_{\cdot}(w)$ on $(0, t]$. However, this definition and the definition of $B(w) = B(\epsilon, t)(w)$ as the set of jump times $s$ on $(0,t]$ such that $\sum_{s \in B(\epsilon, t)(w)} (\Delta X_{s} (w))^{2} \leq \epsilon^{2}$, and $A(\epsilon, t)(w) \cup B(\epsilon, t)(w)$ equals the stopping times of $X_{\cdot}(w)$ on $(0,t]$ does not match. Moreover, the definition of $A = A( \epsilon, t)$ does not depend on $\epsilon $, and $A$ and $B$ are not necessarily disjoint according the definition.

Any comment, idea, or hint would be welcome

  • 1
    $\begingroup$ $A(t,\epsilon)$ does depend on $\epsilon$. Loosely, $B(t,\epsilon)$ is a (small) tail of convergent series, and $A(t,\epsilon)$ is a (finite) partial sum. $\endgroup$ – zhoraster Jul 26 '18 at 5:09
  • $\begingroup$ @zhoraster, thanks for the explanation. $\endgroup$ – Ivan Jul 26 '18 at 14:20
  • $\begingroup$ With your explanation, the proof is very easy to understand. $\endgroup$ – Ivan Jul 26 '18 at 19:42

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