Motivated by the fact the a Brownian motion satisfies the Strong Markov property and by the change of variable formula for Lebesgue-Stieltjes integrals, I am wondering if the following is true.
Let $B$ be a $(\mathscr{F}_t)_{t\geq 0}$-Brownian motion, $H$ a bounded previsible process and $T$ a stopping time with $\mathbb{P}\{T<\infty\}=1$. Does the equality $$ \int_T^{T+t}H_sdB_s=\int_0^tH_{T+s}dB_{T+s} $$ hold true?
So far I have been able to prove that both the LHS and the RHS are well defined $(\mathscr{F}_{T+t})_{t\geq 0}-$martingales, and that the equality holds when $H$ is simple and $T$ assumes only finitely many values. Now I would like to prove it for $H$ simple and arbitrary $T$ by considering $T^{m,n}=2^{-n} ⌈2^n(T\wedge m)+1⌉ $ and then taking limits. But things get problematic and one of the problem is for example that I don't know how to show that the expression $$ \int_0^tH_{T^{m,n}+t}dB_{T+t}-\int_0^tH_{T^{m,n}+t}dB_{T^{m,n}+t} $$ goes to zero.