A question from Stein's Harmonic Analysis - real variable methods' book. In the book of Stein, on page 9:
"Let us remark that these additional properties easily lead to the following conclusions among others. First note that $\mu (B) >0$ for any ball $B$, which is a consequence of the doubling condition, (iii), and the fact that $\mu( \mathbb{R}^n) >0$.
It then follows from (iv) that for any locally integrable $f$ and any $\delta>0$, the mean value:
$$A_\delta(f(x)) = \frac{1}{\mu(B(x,\delta))}\int_{B(x,\delta)} f(y) d\mu(y)$$
is continuous as a function of $x$.
"
The second conclusion of $A_\delta(f(X))$ is what I don't know how to conclude.
Do you have any hints?
Here are the conditions listed:
(i) $B(x,\delta)\cap B(y,\delta) \ne \emptyset $ implies $B(y,\delta)\subset B(x, c_1 \delta)$.
(ii) $\mu(B(x,c_1\delta))\le c_2 \mu(B(x,\delta))$ where $c_1,c_2>1$ are constants.(this is the doubling condition).
(iii) $\bigcap_\delta \overline{B}(x,\delta) = \{ x\} $, $\bigcup_\delta B(x,\delta)=\mathbb{R}^n$.
(iv) For each open set $U$ and each $\delta>0$ the function $x\mapsto \mu(\{ B(x,\delta)\cap U\} ) $ is continuous.
 A: As you pointed out to Felipeh, it's important to remember that the bounded open (with respect to Euclidean topology on $\mathbb{R}^{n}$) sets $B(x,\delta)$ are a priori "balls", not actual balls define by a metric or some weaker quasidistance function. Perhaps, you have already shown this, but I don't think it is that obvious that $\mu$ is finite on the balls $B(x,\delta)$ and inner regular, which Stein uses in the proof of the weak type $(1,1)$ inequality for the associated maximal operator. One can show this, but I thought it was a nontrivial exercise.
For the benefit of those users who don't have access to the book, here's a relevant passage on pg. 8 at the beginning of the section describing what these sets are:

"We shall assume we are given, for each $x\in\mathbb{R}^{n}$, a collection $\left\{B(x,\delta)\right\}$ of nonempty open, bounded subsets of $\mathbb{R}^{n}$, parametrized by $\delta, 0<\delta<\infty$; that is $B=B(x,\delta)$ is the "ball", "centered" at $x$ of "radius" $\delta$. We shall suppose that the balls are monotonic in $\delta$ in the sense that $B(x,\delta_{1})\subset B(x,\delta_{2})$, whenver $\delta_{1}<\delta_{2}$. We shall also assume that we are given a nonnegative Borel measure $\mu$ with the property that $\mu(\mathbb{R}^{n})>0$.
...We assume there exist constants $c_{1}$ and $c_{2}$ both greater than $1$, so that, for all $x$, $y$, and $\delta$..."

The OP has the hypotheses in the question.

Regarding your question, note that condition (iv) implies that 

for fixed $\delta>0$ the function $x\mapsto \mu(B(x,\delta)\cap\mathbb{R}^{n})=\mu(B(x,\delta))$ is continuous.

Now fix $x\in\mathbb{R}^{n}$ and $\delta>0$, and suppose that
$$\mu(B(x,\delta)\triangle B(y,\delta))\not\rightarrow 0, \quad y\rightarrow x$$
Then there exists some $c>0$ and a sequence $y_{k}\rightarrow y$, such that
$$\mu(B(x,\delta)\setminus B(y_{k},\delta))+\mu(B(y_{k},\delta)\setminus B(x,\delta))=\mu(B(x,\delta)\triangle B(y_{k},\delta))\geq c, \quad\forall k$$
Given $0<\epsilon<\min\left\{c/2,\mu(B(x,\delta))/2\right\}$, there exists an open neighborhood $U$ about $x$, such that
$$y\in U\Longrightarrow \left|\mu(B(x,\delta)-\mu(B(y,\delta))\right|<\epsilon$$
Since the function $y\mapsto \mu(B(x,\delta)\cap B(y,\delta))$ is also continuous, replacing $U$ by a smaller open neighborhood of $x$, we have that
$$y\in U\Longrightarrow \mu(B(x,\delta)\setminus B(y,\delta))=\mu(B(x,\delta))-\mu(B(x,\delta)\cap B(y,\delta))<\epsilon,$$
Without loss of generality, we may assume that $y_{k}\in U$ for all $k$, therefore
$$\mu(B(y_{k},\delta)\setminus B(x,\delta))\geq c/2,\quad\forall k$$
and therefore
\begin{align*}
\mu(B(x,\delta))&=\lim_{k\rightarrow\infty}\mu(B(y_{k},\delta))=\mu(B(y_{k},\delta)\setminus B(x,\delta))+\mu(B(y_{k},\delta)\cap B(x,\delta))\\
&\geq c/2+\lim_{k\rightarrow\infty}\mu(B(y_{k},\delta)\cap B(x,\delta))\\
&=c/2+\mu(B(x,\delta)),
\end{align*}
which is a contradiction. Summarizing,

$$\mu(B(x,\delta)\triangle B(y,\delta))\rightarrow 0, y\rightarrow x$$

We'll need the following lemma.

Lemma. If $f$ is locally integrable with respect to $\mu$, then given $\epsilon>0$ and a bounded measurable set $K\subset\mathbb{R}^{n}$, there exists $\gamma>0$, such that
  $$\text{$E\subset K$ measurable, $\mu(E)<\gamma$}\Longrightarrow\int_{E}\left|f(y)\right|d\mu(y)<\epsilon/2$$

Proof. By local integrability, the function $f\chi_{K}\in L_{\mu}^{1}(\mathbb{R}^{n})$. Let $g\in L_{\mu}^{1}(\mathbb{R}^{n})$ be a bounded function such that $\left\|f-g\right\|_{1}<\epsilon$. Then
\begin{align*}
\int_{E}\left|f(y)\right|d\mu(y)&=\int_{E}\left|[f\chi_{K}(y)-g(y)]+g(y)\right|d\mu(y)\\
&\leq\left\|f\chi_{K}-g\right\|_{L_{\mu}^{1}}+\int_{E}\left|g(y)\right|d\mu(y)\\
&\leq\epsilon/2+\left\|g\right\|_{L_{\mu}^{\infty}}\left|E\right|
\end{align*}
Taking $\gamma<\epsilon/(2\left\|g\right\|_{\infty})$ completes the proof. $\Box$
In order to apply the lemma, we need to show that for all $y$ closed to $x$, the union $B(x,\delta)\cup B(y,\delta)$ are contained in some bounded measurable set, which may depend on $x$. But this is trivial, as $U=B(x,\delta)$ is an open neighborhood about $x$, and for all $y\in U$, $B(x,\delta)\cap B(y,\delta)\neq\emptyset$, whence by condition (i),
$$B(x,\delta)\cup B(y,\delta)\subset B(x,c_{1}\delta),$$
which is a bounded, measurable set. As shown above, there exists an open neighborhood $U$ about $x$ such that
$$y\in U\Rightarrow\mu(B(x,\delta)\triangle B(y,\delta))<\gamma\Rightarrow\int_{B(x,\delta)\triangle B(y,\delta)}\left|f(z)\right|d\mu(z)<\epsilon,$$
given $\epsilon>0$. It follows immediately that
$$\lim_{y\rightarrow x}\int_{B(y,\delta)}\left|f(z)\right|d\mu(z)=\int_{B(x,\delta)}\left|f(z)\right|d\mu(z)$$
Putting the above result together with $\mu(B(y,\delta))\rightarrow\mu(B(x,\delta))$, as $y\rightarrow x$, completes the proof.
A: Let us try to show that $A_\delta(f(x))$ is continuous at $x$.  We will want to choose a $\rho > 0$ such that 
$$
|A_\delta(f(x)) - A_\delta(f(y))| < \varepsilon
$$
when $d(x,y)<\rho$.  But we can bound this difference using the triangle inequality:
$$
|A_\delta(f(x)) - A_\delta(f(y))| \leq \int_{B(x,\delta)\Delta B(y,\Delta)} |f(y)|\,d\mu(y),
$$
where $B(x,\delta)\Delta B(y,\delta)$ refers to the symmetric difference of the two balls.  Geometrically, when $x$ and $y$ are close to each other, this symmetric difference is contained in a small shell around the boundary of $B(x,\delta)$.  
For notation, define the shell $S(x,\delta,\rho)$ with width $\rho$ by
$$
S(x,\delta,\rho) := \{z; \delta-\rho \leq d(z,x) \leq \delta+\rho\}.
$$
Now we can state more precisely that when $d(x,y)<\rho$, $B(x,\delta)\Delta B(y,\delta)\subset S(x,\delta,\rho)$,* so that 
$$
|A_\delta(f(x)) - A_\delta(f(y))| \leq \int_{S(x,\delta,\rho)} |f(y)|\,d\mu(y)
$$
Because $f$ is locally integrable, there exists some $\eta>0$ such that 
$$
\int_E |f(y)|\,d\mu(y) < \varepsilon
$$
whenever $E\subset B(x,2\delta)$ and $|E|<\eta$.  This can be proven, for example, by approximating $f$ by bounded functions and using dominated convergence. 
Thus, we are done once we can show that we can choose $\rho$ so small that 
$\mu(S(x,\delta,\rho)) < \eta$.  But observe that
$$
\mu(S(x,\delta,\rho)) = \mu(B(x,\delta+\rho)) - \mu(B(x,\delta-\rho)).
$$
Therefore condition (iv) implies that we can choose $\rho$ so small that 
$\mu(S(x,\delta,\rho)) < \eta$ as desired.
EDIT:
We can prove the claim $B(x,\delta)\Delta B(y,\delta) \subset S(x,\delta,\rho)$
when $d(x,y)<\rho$ using the definition of a metric (not relying on the Euclidean one).
If $z\in B(x,\delta)\Delta B(y,\delta)$ then either $z\in B(x,\delta)\setminus B(y,\delta)$ or $z\in B(y,\delta)\setminus B(x,\delta)$.


*

*Case I: If $z\in B(x,\delta) \setminus B(y,\delta)$, then $d(z,y) > \delta$.  Thus by the triangle inequality
$$
\delta < d(z,y) \leq d(z,x) + d(x,y) < d(z,x) + \rho.
$$
Subtracting $\rho$ from both sides and using $d(z,x) < \delta$ gives
$$
\delta-\rho < d(z,x) < \delta,
$$
which suffices for $z\in S(x,\delta,\rho)$.

*Case II: If $z\in B(y,\delta) \setminus B(x,\delta)$, then by 
$d(z,y) < \delta$ and the triangle inequality,
$$
d(z,x) \leq d(z,y) + d(x,y) \leq \delta + \rho.
$$
Moreover $z\not\in B(x,\delta)$ implies $d(x,z) \geq \delta$, so we conclude that
$$
\delta \leq d(x,z) \leq \delta+\rho
$$
which again suffices for $z\in S(x,\delta,\rho)$.
