I am trying to prove the mean-value property for harmonic functions in $R^k$ by ito calculus.
given $G$ bounded domain and $u$ harmonic function on $G$ then
$u(a)=\int_{\partial B_r} u(y)ds(y)$
$B_r$ is a ball in radius $r$ around a.
so far I took the stopping time $\tau:=\inf_{t\geq0} (t:B_t \notin G ) $ (_hitting time of brownian motion)
and used $\tau_b$ as hitting time of BM in a ball of radius r around a.
I used ito formula (with $B$ brownian motion) to get that
$u(B_{t \wedge \tau_b})=u(B_0)+\sum_{i=1}^{k}\int_{0}^{t \wedge \tau_b} \frac{\partial u}{\partial x_i}(B_s)dB_s+\frac{1}{2}\int_{0}^{t \wedge \tau_b}\Delta u ds $
I know the last term is zero since $u$ harmonic but how can I continue from here I don't see the relation with integral over the ball surface
thank you