Given a $X$ semi-martingale on a filtered probability space $(\Omega,\{\mathcal F_t\}_{t\le\infty},P)$ I am trying to prove:
For any $B\in\mathcal F_\infty$ and processes $a_1,a_2$ such that $\int_t^T a_i\;d\!X$ is well-defined for $i\in\{1,2\}$ it holds:
$$a_1=a_2 \text{ on } B\Longrightarrow P\left(\int_t^T a_1\;d\!X=\int_t^T a_2\;d\!X\middle|B\right)=1$$
My proof
Let $(a_i^n)_{n<\infty}$ be a series of simple integrands approximating $a_i$. We can choose them such that $a_1^n=a_2^n$ on $B$ for all $n$. Define $A^n_i=\int_t^T a^n_i\;d\! X$. These integrals of simple integrands are defined path-wise and thus $A_1^n=A_2^n$ on $B$ for each $n$. The integrals of $a_i$ are given by $p\!\lim_{n\rightarrow \infty} A^n_i$ (limit in probability). Convergence in probability preserves the equivalence on $B$ and the proof it complete.
Question
It the proof correct and complete? Is there an easier/correct proof?
Thx