I have a question based on book "Markov processes, Brownian motion and time symmetry", on page 113, proposition 10 says, under Hypothesis(L), for any nearly borel set A, there is a sequence of compact sets $(K_n)\uparrow$ and does not depent on $x$ such that $P^x\{\lim_{n\rightarrow \infty} T_{K_n}=T_A\}=1$, where $T$ denotes the hitting time for the sets. Hypothesis(L) is given on page112 and followed by an argument that (seems to me) Brownian motion satisfies this hypothesis(L).
Now, if we focus on Brownian motion, by above statement, $E^x(e^{-T_{K_n}})\rightarrow E^x(e^{-T_A})$ point-wisely. With $x\rightarrow E^x(e^{-T_{K_n}})$ are all borel measurable, one can conclude $x\rightarrow E^x(e^{-T_{A}})$ is borel measurable. But $E^x(e^{-T_{A}})=1$ iff $x$ is regular to $A$, so we have: for a nearly Borel set $A$, the regular set $A^r$ is Borel?