Let $\mathbb{F} \subset \mathbb{G}$ where $\mathbb{G}$ is the filtration progressively enlarged by the filtration $\mathbb{F}$ and let $\tau$ be a $\mathbb{G}$-stopping time.
Define $B_t:= \exp (\int_0^t r_s ds)$ where $r$ is $\mathbb{G}$-adapted and let $A$ be a $\mathbb{G}$-adapted process with finite variation, can some one explain the following equality?
$$B_\tau\mathbb{E} \bigg[\int_{[\tau,T]} (B_s)^{-1} dA_s \bigg| \mathcal{G}_\tau \bigg]= A_\tau - A_{\tau-} + B_\tau\mathbb{E} \bigg[\int_{(\tau,T]} (B_s)^{-1} dA_s \bigg| \mathcal{G}_\tau \bigg]$$
How does the right hand side change if $r$ and\or $A$ were $\mathbb{F}$-adapted?
Since the integral $\int_{\mathbb{R}_+}B_s^{-1}\,dA_s$ is a (stochastic) integral by path, it is also a Lebesgue-Stieljes type integral, then \begin{equation*} \int_{[a,b]}B_s^{-1}\,dA_s = \int_{\mathbb{R}_+}1_{[a,b]}(s)B_s^{-1}\,dA_s,\quad 0\le a\le b<\infty, \end{equation*} Meanwhile, $1_{[\tau,T]}(s)=1_{[\tau,\tau]}(s)+1_{(\tau,T]}(s) $, then \begin{align*} \int_{[\tau,T]}B_s^{-1}\,dA_s&=\int_{\mathbb{R}_+}1_{[\tau,T]}(s)B_s^{-1}\,dA_s\\ &=\int_{\mathbb{R}_+}1_{[\tau,\tau]}(s)B_s^{-1}\,dA_s +\int_{\mathbb{R}_+}1_{(\tau,T]}(s)B_s^{-1}\,dA_s\\ &=B_{\tau}^{-1}\Delta A_{\tau}+\int_{(\tau,T]}B_s^{-1}\,dA_s. \end{align*} Therefore, \begin{equation*} B_\tau \mathsf{E}[\int_{[\tau,T]}B_s^{-1}\,dA_s| \mathcal{G}_\tau] =B_\tau \mathsf{E}[B_{\tau}^{-1}\Delta A_{\tau}| \mathcal{G}_\tau ] + B_\tau \mathsf{E}[\int_{(\tau,T]}B_s^{-1}\,dA_s| \mathcal{G}_\tau] \end{equation*}