Consider disjoint sets A, B and Brownian motion $B_{t}$ with $B_{0}\notin A\cup B$.
Let $T_{A}:=inf_{t>0}\{B_{t}\in A\}$. Then, do we get $P(T_{A\cup B}<\infty)=P(T_{A}<\infty)+P(T_{ B}<\infty)-P(T_{A\cap B}<\infty)=P(T_{A}<\infty)+P(T_{ B}<\infty)+0$?
I think so because $( T_{ A}<\infty )\cap (T_{ B}<\infty)=inf_{t}(B_{t}\in A\cap B)$.
Thnx
No!
The event $T_{A \cup B} < \infty$ means the Brownian motion eventually hits $A \cup B$. This is the union of $T_A < \infty$ and $T_B < \infty$, so $$ P(T_{A \cup B} < \infty) = P(T_A < \infty) + P(T_B < \infty) - P((T_A < \infty) \cap (T_B < \infty))$$ But $(T_A < \infty) \cap (T_B < \infty)$ is not the same as $(T_{A \cap B} < \infty)$: if it hits $A$ and hits $B$, there's no reason why it should hit the intersection: the times when it hits $A$ can be different from the times when it hits $B$. Indeed, in this particular case you're assuming $A$ and $B$ are disjoint, so there is no intersection. But it's quite possible that both $P(T_A < \infty) = 1$ and $P(T_B < \infty) = 1$, and then of course $P((T_A < \infty) \cap (T_B < \infty)) = 1$.