I have a slight confusion about the definition of independence under a conditional probability measure. So the goal is to show that given $P(T<\infty)>0$, and $B_t^{(T)}=1_{\{T<\infty\}}(B_{T+t}-B_T),$ then under the probability measure $P(\cdot | T<\infty)$, the process $(B_t^{(T)})_{t\ge 0}$ is a Brownian motion independent of $\mathscr{F}_T$.
The proof is completed by showing that when $P(\{T=\infty\})>0$,
$$E[1_{A\cap \{T<\infty\}}F(B_{t_1}^{(T)},\dots, B_{t_p}^{(T)})]=P(A\cap \{T<\infty\})E[F(B_{t_1},\dots, B_{t_p})].$$
I am not clear why this proves the independence of the finite dimensional distributions with $\mathscr{F}_T$ under the conditional probability measure $P(\cdot|T<\infty)$. From my knowledge this measure is given by $P(A|T<\infty)= P(A\cap \{T<\infty\})/P(\{T<\infty\})$. So don't we need to show $$E[1_{A\cap \{T<\infty\}}F(B_{t_1}^{(T)},\dots, B_{t_p}^{(T)})]=P(A\cap \{T<\infty\})E[1_{\{T<\infty\}}F(B^{(T)}_{t_1},\dots, B^{(T)}_{t_p})]$$ instead?
Also, I think $E[1_{ \{T<\infty\}}F(B_{t_1}^{(T)},\dots, B_{t_p}^{(T)})]=P(\{T<\infty\})E[F(B_{t_1},\dots, B_{t_p})]$ (the case with $A=\Omega$) implies that $(B_{t_1}^{(T)},\dots, B_{t_p}^{(T)})$ under $P(\cdot | T<\infty)$ has the same distribution as $(B_{t_1},\dots , B_{t_p})$ because we can divide both sides by $P(T<\infty)$ then the equality implies that the conditional probability of $(B_{t_1}^{(T)},\dots, B_{t_p}^{(T)})$ equals the distribution of $(B_{t_1},\dots, B_{t_p})$. Is this reasoning correct?
If this is correct, then I think the conditional independence can be proved by dividing the first equality by $P(T<\infty)$ so $$E[1_{A\cap \{T<\infty\}}F(B_{t_1}^{(T)},\dots, B_{t_p}^{(T)})]/P(T<\infty)=P(A\cap \{T<\infty\})/P(T<\infty)\cdot E[F(B_{t_1},\dots, B_{t_p})].$$ Then we use that $E[1_{ \{T<\infty\}}F(B_{t_1}^{(T)},\dots, B_{t_p}^{(T)})]=P(\{T<\infty\})E[F(B_{t_1},\dots, B_{t_p})]$ so if we divide and multiply the RHS above by $ P(T<\infty) $ then we get $$E[1_{A\cap \{T<\infty\}}F(B_{t_1}^{(T)},\dots, B_{t_p}^{(T)})]/P(T<\infty)=P(A\cap \{T<\infty\})/P(T<\infty)\cdot E[1_{ \{T<\infty\}}F(B_{t_1}^{(T)},\dots, B_{t_p}^{(T)})]/P(T<\infty).$$
So taking the limits for $F$ that gives the indicator function for cubes, we get conditional independence in the usual form as $P(A\cap B|C)=P(A|C)\cdot P(B|C)$.
Perhaps I am misinterpreting this whole thing. Could anyone clarify what it means that the process $(B_t^{(T)})_{t\ge 0}$ is independent of $\mathscr{F}_T$ under $P(\cdot | T<\infty)$?