Let $X$ be a semimartingale and $V$ be a finite variation process. Lemma 3 of this page of almostsuremath proves that
$$[X,V]_t=\int_0^t\Delta X_s\,dV_s=\sum_{s\leq t}\Delta X_s\,\Delta V_s$$
This is the proof (let $t_{k,n}=k\,t/n$ denote the $k$-th point of the equispaced partition of $[0,t]$)
$$ \begin{eqnarray} [X,V]_t & = & \lim_{n\rightarrow\infty}\sum_{k=1}^n(X_{t_{k,n}}-X_{t_{k-1,n}})\,(V_{t_{k,n}}-V_{t_{k-1,n}}) \nonumber\\ &=&\lim_{n\rightarrow\infty}\sum_{k=1}^n\int_0^t1_{\{t_{k-1,n}\leq s\leq t_{k,n}\}}(X_{t_{k,n}}-X_{t_{k-1,n}})\,dV_s\nonumber\\ &=&\int_0^t\lim_{n\rightarrow\infty}\sum_{k=1}^n1_{\{t_{k-1,n}\leq s\leq t_{k,n}\}}(X_{t_{k,n}}-X_{t_{k-1,n}})\,dV_s\nonumber\\ &=&\int_0^t\Delta X_s\,dV_s \end{eqnarray} $$
It is said that the third equality follows from bounded convergence theorem (which allows to exchange the limit and the integral). I cannot fully understand why. First, I guess that this is the step in which we use the fact that V has finite variation (I do not see any other place). My other guess is that the FV property of $V$ allows to use the bounded convergence theorem. My question is: how are we exactly using the bounded convergence theorem?
The integral is with respect to $dV_s$, so I a random measure and this troubles me even more (I have never applied the theorem in this situation).