Show that $\frac1{|\partial B_r(x)|}\int_{\partial B_r(x)}f\xrightarrow{r\to0+}f(x)$ Let $d\in\mathbb N$ and $\sigma_M$ denote the surface measure on $\mathcal B(M)$ for every embedded $C^1$-submanifold $M$ of $\mathbb R^d$. Moreover, let $$\theta_r:\mathbb R^d\to\mathbb R^d\;,\;\;\;x\mapsto rx$$ for $r>0$.
We have the following result from geometric measure theory:

Theorem 1: If $k\in\{1,\ldots,d\}$, $M$ is a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ and $r>0$, then $\theta_r(M)$ is a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$. Moreover, $f:\theta_r(M)\to\mathbb R$ is $\sigma_{\theta_r(M)}$-integrable iff $f\circ\theta_r$ is $\sigma_M$ integrable. In either case, $$\sigma_{\theta_r(M)}f=r^k\theta_r(\sigma_M)f=r^k\sigma_M(f\circ\theta_r)\tag1,$$ where $\mu g:=\int g\:{\rm d}\mu$ and $\theta_r(\sigma_M)$ denotes the pushforward of $\sigma_M$ wrt $\theta_r$.
Remark 2: If $\sigma_M(M)\in(0,\infty)$, we may normalize $\sigma_M$: $$\alpha_M:=\frac{\sigma_M}{\sigma_M(M)}.$$ Note that $$\alpha_{\theta_r(M)}=\frac{\theta_r(\sigma_M)}{\sigma_M(M)}\tag2.$$

Question: How can we show that if $f:\mathbb R^d\to\mathbb R^d$ is continuous, then $$\alpha_{\partial B_r(x)}f\xrightarrow{r\to0+}f(x)\tag3$$ for all $x\in\mathbb R^d$?
By $(2)$, $$\alpha_{\partial B_r(x)}=\frac{\theta_r(\sigma_{\partial B_1(x)})}{\sigma_{\partial B_1(x)}(\partial B_1(x))}.\tag4$$ So it is enough to consider $$\theta_r(\sigma_{\partial B_1(x)})f=\sigma_{\partial B_1(x)}(f\circ\theta_r)=\int\sigma_{\partial B_1(x)}({\rm d}y)f(ry)\tag5.$$

At this stage it seems like I'm missing something, since the right-hand side of $(5)$ clearly tends to $f(0)\sigma_{\partial B_1(x)}(\partial B_1(x))$ as $t\to0+$. What's missing here seems to be a translation of $f$ by $x$ in the integral on the right-hand side of $(5)$, but where should this translation come from (or, phrased differently, which of my identities above is wrong)?

 A: Let me use $S_r(x):=\partial B_r(x)$ to denote the sphere of radius $r$ centered at $x$. The issue is that if $x\neq 0$, then $S_r(x)$ is NOT the $r$-scaling of $S_1(x)$. Think about it: if $x\neq 0$, and $r>0$ is very small, then $S_r(x)$ is a very small sphere centered at $x$; in particular this means it is "away from the origin". If however you take the unit sphere at $x$, $S_1(x)$, and then scale it by a very small $r>0$ to get $r\cdot S_1(x)$, then the result will be something with very small magnitude, i.e close to the origin.
So, it is not a simple homogeneous transformation, rather the correct relationship is an affine one: $S_r(x)$ is obtained by first scaling $S_1(0)$ by a factor of $r$ to get $S_r(0)$, and then translating by $x$; this is what you were missing. So,
\begin{align}
\left|\frac{1}{\sigma(S_r(x))}\int_{S_r(x)}f(y)\,d\sigma(y) - f(x)\right| &\leq
\frac{1}{\sigma(S_r(x))}\int_{S_r(x)}|f(y)-f(x)|\,d\sigma(y)\\
&=\frac{1}{\sigma(S_1(0))\cdot r^{d-1}}\int_{S_1(0)}|f(x+rz)-f(x)|\cdot r^{d-1}\,d\sigma(z)\\
&=\frac{1}{\sigma(S_1(0))}\int_{S_1(0)}|f(x+rz)-f(x)|\,d\sigma(z),
\end{align}
where I have used the same notation $\sigma$ throughout because context clearly tells us which meaning is intended. From this final expression, continuity of $f$ at $x$ gives the desired conclusion that the limit as $r\to 0^+$ is $0$.
