Yesterday I read Stein's Harmonic Analysis and came across a quite general definition of the maximal function $Mf$ and the uncentered version $\tilde M f$. It is not hard to show the equivalence between the two definition on $\Bbb R^n$ with the usual Lebesgue measure and standard definition of balls. However, the definition I found in Stein's book is a lot more general.
We have a positive Borel measure $\mu$ on $\Bbb R^n$. For each $x\in \Bbb R^n$, a family of balls $\{B(x,\delta)\}_{\delta>0}$ is associated. Each ball is defined to be a bounded, open set such that the family is monotone in the sense that $B(x,\delta_1)\subset B(x,\delta_2)$ whenever $\delta_1<\delta_2$. It is also postulated that $\cup_{\delta>0} B(x,\delta) = \Bbb R^n$ and $\cap_{\delta>0} B(x,\delta) = \{x\}$. The relationship between balls with different centers is given by
There exists $c_1>1$ such that for any $x,y \in\Bbb R^n$ and $\delta>0$, we have $$ B(x,\delta) \cap B(y,\delta) \ne \emptyset \implies B(y,\delta) \subset B(x,c_1\delta). $$ Moreover, for the same $c_1$ as above, there exists $c_2>1$ such that for any $x\in\Bbb R^n$ and $\delta>0$, $$ \mu(B(x,c_1\delta)) \le c_2 \mu(B(x,\delta)). $$
Now, for any $f\in L^1_{\text{loc}}(\Bbb R^n, d\mu)$, the maximal function $Mf$ is defined with respect to such a family of "balls", i.e.
$$
Mf(x) := \sup_{\delta>0} \frac{1}{\mu(B(x,\delta))} \int_{B(x,\delta)} |f| \,d\mu
$$
whereas the uncentered maximal function $\tilde Mf$ is defined analogously, i.e.
$$
\tilde Mf(x) := \sup \left\{ \frac{1}{\mu(B(y,\delta))} \int_{B(y,\delta)} |f| \,d\mu \ :\ x \in B(y,\delta) \right\}.
$$
It is not hard to prove that $$ \tilde Mf \le C \cdot Mf $$ for some $C>0$ under the above assumptions, hence the 2 versions of maximal functions are equivalent. However, it is claimed in the book (without a proof) that this inequality holds under an even weaker assumption on the relationship between $B(x,\delta), B(y,\delta)$. Instead of postulating the existence of $c_1,c_2>1$ as above we instead assume
For each $B(x,\delta)$, we define $B^*(x,\delta)$ to be the set $$ B^*(x,\delta) := \bigcup \left\{ B(y,\delta) : B(x,\delta) \cap B(y,\delta) \ne \emptyset \right\}. $$ There exists $c_3>1$ such that $$ \mu(B^*(x,\delta)) \le c_3\mu(B(x,\delta)). $$
Under this weaker assumption, I can't see how to deduce the inequality $\tilde Mf \le C \cdot Mf$. For any $B(y,\delta)$ containing $x$, $B(y,\delta) \subset B^*(x,\delta)$ so we have the inequality $$ \frac{1}{\mu(B(y,\delta))} \int_{B(y,\delta)} |f| \,d\mu \le \frac{\mu(B^*(x,\delta))}{\mu(B(y,\delta))} \cdot \frac{1}{\mu(B^*(x,\delta))} \int_{B^*(x,\delta)} |f| \,d\mu $$ but since $B^*(x,\delta)$ does not belong to the family of "balls" $\{B(x,\delta)\}_{\delta>0}$, I'm not sure how to bound the averaging integral on the righthand-side (the ratio $\frac{\mu(B^*(x,\delta))}{\mu(B(y,\delta))}$ can be bounded by $(c_3)^2$ though).
Maybe I'm not seeing something obvious here? Could anyone help me figure out what did I miss? The book stated as if the statement is obvious but I can't seem to see why.