# Question about Protter's proof of the Ito's formula

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

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

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

• I'll give it a shot. Let $C(\omega)$ be set of jump times on $[0,t)$. Note that C is countable a.s. by cadlag property (real analysis result). A is not an arbitrary (finite) set, but C\B. It should be clear that B depends on $\epsilon, t$ and therefore A does as well. The only nontrivial question is whether A is finite. Since C is countable and $\sum_{0\leq s <t} (\Delta X_s)^2$ converges, for every $\epsilon$, $\exists N$ s.d. $\sum_{n> N} (\Delta X_{s_n})^2 \leq \epsilon^2$. Choose the smallest $N$ then A contains precisely the first $s_1,\dots, s_{N}$ and B contains the rest. – James Yang Jul 28 '18 at 3:46