$\newcommand{\avint}{⨍}$
Let $u:\mathbb{R}^n \rightarrow \mathbb{R} \quad u\in C^1 \quad r>0$
How do I rigorously understand the limit of the integral $$ \lim_{r\rightarrow0}\frac{1}{nw_nr^{n-1}} \int_{\partial B(x,r)}u(y) dy=\lim_{r\rightarrow0}\avint_{\partial B(x,r)} u(y)dy=u(x)$$ where $w_n $ is the volume of the UNIT ball in n dimensions.So the denominator is actually the surface area of the ball.
This looks very intuitive but how do I understand it rigorously.
All you need is continuity of $u$ at $x$. Let $\epsilon >0$. If $r$ is sufficiently small and $y \in B(x,r)$ then $|u(y)-u(x)|<\epsilon$. Observe that $nw_nr^{n-1}$ is the measure of the sphere $\partial B(x,r)$. Let us denote this by $a_n$. Then $|\frac 1 {a_n} \int_{\partial B(x,r)} u(y)dy -u(x)|\leq \frac 1 {a_n} \int_{\partial B(x,r)} |u(y)-u(x)|dy <\epsilon$. [Some basic fact used here are: the integral over $\partial B(x,r)$ is with respect to the 'uniform measure' on the sphere; the measure of $\partial B(x,r)$ under this measure equals $nw_nr^{n-1}$].