proof of the mean value property in Evan's PDE book

Question

I understood everything but the last part which I highlighted in green. as $t \to 0$ I can only imagine that $B(x,t)$ becomes the point $x$ and so does $\partial B(x,t) $ because I think the boundary of the point is the point itself, right ? so the integral over a point is just computing the value of the function at that point ?

sorry if I said any non-sense but I'd like a clear explanation of that last step. thanks !

Answer 1

Note that

$$\tag 1\int_{\partial B(x,t)}u(y)\,dS(y) - u(x) = \int_{\partial B(x,t)}(u(y)-u(x))\,dS(y).$$

Now slap absolute values on and use the continuity of $u$ at $x$ to see

$$ \sup_{y\in \partial B(x,t)}|u(y)-u(x)| \to 0 \text { as } t \to 0.$$

This shows $(1) \to 0$ as desired.

Answer 2

This is a property of absolutely continuous functions (see Lebesgue's differentiation theorem). Since $u \in C^2 (U)$, we must have that the limit taken in the problem is simply $u(x)$.