let $D$ open set of $\mathbb{R}^{n}$ and $T_{D}=\inf\{t\geq 0 : X_{t}\notin D\} $ be the first exit time from the $D$ and $1_{A}$ is Indicator function of $A \subseteq \partial D$ $$ \text{Calculated and interpreted } E^{x}[1_{A}(X_{T_{D}})] $$
Indeed: $$ E^{x}[1_{A}(X_{T_{D}})]=E^{x}[1_{\{X_{T_{D}}\in A\}}]=P^{x}[X_{T_{D}}\in A] $$ Can we say that $P^{x}[X_{T_{D}}\in A]$ is equal the probability of exit D from A
Please respond
Let $\omega \in \Omega$. By definition of the stopping time, $X_{T_D}(\omega)$ tells us at which point the sample path $t \mapsto X_{t}(\omega)$ leaves for the first time the set $D$. Consequently, $1_A(X_{T_D})(\omega)$ equals $1$ if and only if this point is an element of $A$. Therefore $\mathbb{P}^x[X_{T_D} \in A]$ is the probability that the point of exit (from $D$) is contained in $A$, if the process is started at $x \in \mathbb{R}^n$.
Example Let $(B_t)_{t \geq 0}$ a one-dimensional Brownian motion, $B_0=0$, and define $$T:= T_{(-a,b)} := \inf\{t \geq 0; B_t \notin (-a,b)\}$$ for $a,b>0$. Obviously, $(B_t)_{t \geq 0}$ exits from the interval $(-a,b)$ either in $-a$ or in $b$ and one can show that
$$\mathbb{P}(B_{T_D} \in \{-a\}) \stackrel{\text{def}}{=} \mathbb{P}(B_T = -a) = \frac{b}{a+b} \qquad \mathbb{P}(B_{T_D} \in \{b\}) \stackrel{\text{def}}{=} \mathbb{P}(B_T = b) = \frac{a}{a+b}$$