Expectation of a function of Brownian motion at a stopping time I do not understand the following claim from my book:

Let $(B_t)$ be a Brownian motion on $\mathbb{R}^d$ starting at $x$.  Let $\tau = \inf \{ t>0 : B_t \in \partial B( x, r) \}$. Also, let $u : \mathbb{R}^d \rightarrow \mathbb{R}$ be some integrable function. How can we show that
  $$ \mathbb{E} [ u(B_{\tau}) ] = \frac{1}{\sigma_{x,r} (\partial B(x,r))} \int_{\partial B(x,r)} u(y) \,d \sigma_{x,r} (y) \quad ?$$

Here $\sigma_{x,r}$ denotes the surface area measure on $\partial B(x,r)$. I know that intuitively the expectation is just the average of all the values on the boundary of $B(x,r)$, and hence this expression is obvious. But how can we prove it rigorously?
 A: Rotational invariance tells you that $B_\tau$ has the uniform distribution on the sphere. So the expectation of $u(B_\tau)$ is the integral of $u(x)$ against the uniform distribution on the sphere. This is precisely the integral of $u$ over the sphere rescaled by the surface area, which is your expression.
Now we can talk about proving rotational invariance. Ultimately the easiest argument comes down to checking that if $B_t$ is a Brownian motion and $Q$ is an orthogonal matrix then 
$$(Q B_s)^T (Q B_t) = B_s^T Q^T Q B_t = B_s^T B_t.$$
Hence $\mathbb{E} (Q B_s)^T (Q B_t) = \min \{ s,t \}$. $Q B_t$ clearly has a.s. continuous paths and $Q B_0 = Q 0 = 0$. From there it is standard to prove that $Q B_t$ is also a Brownian motion under whatever definition you use for Brownian motion. (In my class we took this as the definition from the start; some prefer to start from independent increments.)
The above doesn't depend on which orthogonal matrix you choose for $Q$. This means $B_\tau$ has the same distribution as $Q B_\tau$ for any orthogonal matrix $Q$. The only distribution on the sphere with this property is the uniform distribution. 
