The statement looks slighlty incorrect (in my opinion, it leads one to think that $A_0$ must be somewhat special). There is a more general result in real analysis which says
Lemma. Let $\Omega \subset \mathbb R^n$ be open, $x \in \Omega$, $\rho < \mbox{dist}\,(x, \partial \Omega)$. Let $g \geq 0$ be integrable on $B_\rho(x)$. Then for each $\theta \in (0,1)$ it holds
\begin{equation}\tag 1\int_{\partial B_\sigma(x)} g \leq 2 \theta^{-1}\rho^{-1} \int_{B_\rho(x)\setminus B_{\rho/2}(x)} g \end{equation}
for all $\sigma \in (\rho/2, \rho)$ except for a set of measure at most $\theta \rho / 2$.
Proof. Recall the general identity (which follows by Fubini's Theorem and that is valid for each $L^1$ function $g$)
\begin{equation}\tag 2 \int_{\rho/2}^\rho \int_{\partial B_\sigma(x)} g = \int_{B_\rho(x)\setminus B_{\rho/2}(x)} g \end{equation}
Now suppose, for the sake of a contraddiction, that there is a set $\Sigma \subset (\rho/2,\rho)$, with measure $> \theta\rho/2$, on which it holds
\begin{equation}\tag 3 \int_{\partial B_\sigma(x)} g > 2 \theta^{-1}\theta^{-1} \int_{B_\rho(x)\setminus B_{\rho/2}(x)} g \end{equation}
Set $I := (\rho/2, \rho)$ and decompose $I = (I\setminus \Sigma) \cup \Sigma$. By (2),
\begin{equation}\tag 4\int_{\rho/2}^\rho \int_{\partial B_\sigma(x)} g = \int_{I\setminus \Sigma} \int_{\partial B_\sigma(x)} g + \int_{\Sigma} \int_{\partial B_\sigma(x)} g = \int_{B_\rho(x)\setminus B_{\rho/2}(x)} g \end{equation}
The first addend in the middle term is nonnegative, while the second is greater than $|\Sigma| 2\theta^{-1} \rho{-1} \int_{B_\rho(x)\setminus B_{\rho/2}(x)} g$. But if $|\Sigma| > \theta \rho /2$, then the prefactor is strictly greater than 1, hence (4) is absurd. $\Box$
By taking $\theta = 1/4$ and $\rho= 1$, it is now clear that the inequality in question holds for all $A_0$ and for all those $r \in (1/2, 1) \setminus \Sigma$, where $|\Sigma| \leq 1/8$. Hence, an $r \in (1/2, 1)$ with the required property surely exists.
