Let $U\subseteq\mathbb R^d$ be a bounded domain.
A continuous function $u:\overline U\to\mathbb R$ is called harmonic $:\Leftrightarrow$ $$u(x)=\frac 1{|\partial B_r(x)|}\int_{\partial B_r(x)} u\;do\;\;\;\text{for any open ball }B_r(x)\subset\subset U\;.$$
Now, let
- $(\Omega,\mathcal A)$ be a measurable space
- $\mathbb F=\left(\mathcal F_t\right)_{t\ge 0}$ be a filtration on $(\Omega,\mathcal A)$
- $\left(\operatorname P_x\right)_{x\in\mathbb R^d}$ be a family of probability measures on $(\Omega,\mathcal A)$
- $(B_t-x)_{t\ge 0}$ be a $d$-dimensional standard Brownian motion on $(\Omega,\mathcal A,\operatorname P_x)$ with respect to $\mathbb F$, for all $x\in\mathbb R^d$
Let $$\tau:=\inf\left\{t\ge 0:B_t\in\partial U\right\}$$ and $\varphi:\partial U\to\mathbb R$ be Borel-measurable. How can we show, that $$u(x):=\operatorname E_x\left[1_{\left\{\tau<\infty\right\}}\varphi\left(B_\tau\right)\right]\;\;\;\text{for }x\in U$$ is harmonic?
Let $B_r(x)\subset\subset U$ and $$\sigma:=\inf\left\{t>0:B_t\notin B_r(x)\right\}\;.$$ The strong Markov property implies, that \begin{equation} \begin{split} u(x)&=\operatorname E_x\left[\operatorname E_x\left[1_{\left\{\tau<\infty\right\}}\varphi\left(B_\tau\right)\mid\mathcal F_{\sigma}\right]\right]\\ &=\operatorname E_x\left[u\left(B_{\sigma}\right]\right] \end{split} \end{equation}
for all $x\in U$. In Brownian Motion by Peter Mörters and Yuval Peres they state in Theorem 3.8, that the last expression would be equal to $$\int_{\partial B_r(x)}u\;d\varpi\;,$$ where $\varpi$ denotes the uniform distribution on $\partial B_r(x)$. I absolutely don't understand why this is true and why this yields the desired statement.