Assume that $d \geq 2$ and fix any $a \in D$. We would like to prove the following from theorem 2.21 Karatzas and Shreve p245 :
Theorem: $$\lim_{{x \to a, x \in D}} E^x \{ f(W_{\tau_D}) \} = f(a)$$ holds for every bounded, measurable function $f: \partial D \to \mathbb{R}$ which is continuous at $a$ $\implies$ $a$ is regular for $D$.
The following definitions are in play:
2.9 Definition. Consider the stopping time of the right-continuous filtration $\{ \mathcal{F}_t \}$ given by $$ \sigma_D \triangleq \inf\{ t \geq 0 : W_t \in D^c \} $$ (contrast with the definition of $\tau_D$ in (2.1)). We say that a point $a \in \partial D$ is regular for $D$ if $P^a[\sigma_D = 0] = 1$; i.e., a Brownian path started at $a$ does not immediately return to $D$ and remain there for a nonempty time interval.
2.10 Remark. A point $a \in \partial D$ is called irregular if $P^a[\sigma_D = 0] < 1$; however, the event $\{\sigma_D = 0\}$ belongs to $\mathcal{F}^W_{0+}$, and so the Blumenthal zero-one law (Theorem 2.7.17) gives for an irregular point $a$: $P^a[\sigma_D = 0] = 0$.
This is the proof for theorem given. Proof: We assume without loss of generality that $a = 0$, and attempt to prove the theorem part by contradiction. If the origin is irregular, then $P^0[\tau_0 = 0] = 0$ (Remark 2.10). Since a Brownian motion of dimension $d \geq 2$ never returns to its starting point (Proposition 3.3.22), we have $$ \lim_{{r \to 0}} P^0[W_{\sigma_D} \in B_r] = P^0[W_{\sigma_D} = 0] = 0. $$
Fix $r > 0$ for which $P^0[W_{\sigma_D} \in B_r]< 1/4$, and choose a sequence $\{\delta_n\}$ for which $0 < \delta_n < r$ for all $n$ and $\delta_n \to 0$. With $\tau_n = \inf \{ t \geq 0; \|W_t\| \geq \delta_n \}$, we have $P[\tau_n \leq 0] = 1$, and thus, $\lim_{n \to \infty} P^0[\tau_n < \sigma_D] = 1$. Furthermore, on the event $\{\tau_n < \sigma_D\}$ we have $W_{\tau_n} \in D$. For $n$ large enough so that $P^0[\tau_n < \sigma_D] \geq 1/2$, we may write
\begin{aligned} \frac{1}{4} &> P^0[W_{\sigma_D} \in B_r] \\ &\geq P^0[W_{\sigma_D} \in B_r, \tau_n < \sigma_D] \\ \text{(*)} &= E^0(1_{\tau_n<\sigma_D} P^0[W_{\sigma_D} \in B_r|\mathcal{F}_{\tau_n}]) \\ \text{(**)}&= \int_{D \cap B_{\delta_n}} P^x[W_{\tau_D} \geq B_r]P^0[\tau_n < \sigma_D, W_{\tau_n} \in dx] \\ &\geq \frac{1}{2} \inf_{x \in D \cap B_{\delta_n}} P^x[W_{\tau_B} \geq B], \end{aligned}
from which we conclude that $P^x[W_{\tau_B} \geq B] \leq 1/2$ for some $x \in D \cap B_{\delta_n}$. Now choose a bounded, continuous function $f: \partial D \to \mathbb{R}$ such that $f = 0$ outside $B$, $f \leq 1$ inside $B$, and $f(0) = 1$. For such a function we have $$ \overline{\lim}_{n \to \infty} E^x[f(W_{\tau_D})] \leq \overline{\lim}_{n \to \infty} P^x[W_{\tau_D} \in B_r] \leq \frac{1}{2} f(0), $$ and $\lim_{x \to a, x \in D} E^x \{ W_{\tau_D} \} = f(a)$ fails.
QED
My Question:
I have shown explicitly using tower property and take out what is known property of conditional expectation,(*). This is done below. $$P^0[W_{\sigma_D} \in B_r, \tau_n < \sigma_D]= E^0[1_{W_{\sigma_D} \in B_r}1_{\tau_n < \sigma_D}]= E^0[E^0[1_{W_{\sigma_D} \in B_r}1_{\tau_n < \sigma_D}|\mathcal{F}_{\tau_n}]]= E^0[1_{\tau_n < \sigma_D}E^0[1_{W_{\sigma_D} \in B_r}|\mathcal{F}_{\tau_n}]]=E^0(1_{\tau_n<\sigma_D} P^0[W_{\sigma_D} \in B_r|\mathcal{F}_{\tau_n}])$$
How do I go from this, to (**). The chapter has not every represented a probability measure like this before. I'm not looking for intuitive answers, more explicit rigorous reason justifying this step. But intuition as to why would be still helpful in eventually making something rigorous so its still invited as an answer.