If $w$ is in weak $A_{\infty}(d\mu)$ where $d\mu$ is a doubling measure, then is $w\,d\mu$ doubling? Let $\mu$ be a positive Borel measure on $\mathbb{R}^n$ and let it be doubling i.e. there exists a a constant $C>1$ such that $\mu(B(x_0, 2r)) \leq C \mu(B(x_0,r))$ for all balls $B(x_0,r)$. Let $w$ be a weight with respect to $\mu$ i.e. $w\in L^1_\text{loc}(\mathbb{R^n}, d\mu)$ and $w>0$ $\mu$ a.e. Consider the measure $w\,d\mu$ and let
$$w(A) = \int_A w\,d\mu $$ for any measurable set $A$. Suppose there exists constants $\alpha$ and $\beta$ with $0<\alpha, \beta <1$ such that whenever $E\subset B$ is a measurable subset of the ball $B$ and satisfies
$$ \frac{\mu(E)}{\mu(B)} >\alpha  \quad \text{ then} \quad \frac{w(E)}{w(B)} > \beta$$
Question: Is the measure $wd\mu$ doubling? i.e. does there a constant $C>1$ such that $w(B(x_0,2r))\leq Cw(B(x_0,r))$ for all balls $B(x_0,r)$?
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Background: This is the characterization of weak $A_\infty$ weights of $d\mu$. The thing is that if $\mu$ is the lebesgue measure then there is no problem. However for general doubling measures $\mu$, this looks tricky to me as we have no control over the value of $\alpha$ (Note in particular that $\mu(B)/\mu(2B)$ can be quite small as compared to $\alpha$, which makes the given property difficult to use). 
Stein in his Harmonic analysis only has $\mu$ as the lebesgue measure and Journe in book on Calderon Zygmund Operators, says this is obvious but I am skeptical. The best that I could do was that $w$ has to satisfy a reverse holder inequality (just adapting the proof in Stein), but still with this I don't know how to prove doubling. 
 A: You probably know that the doubling condition can be equivalently stated with any other number $>1$ in place of $2$. Indeed, the inequality $\mu(\lambda B)\le C\mu(B)$ implies $\mu(\lambda^kB)\le C^k \mu(B)$ for all positive integers $k$. So, we can upgrade from any scaling factor $\lambda>1$ to an arbitrarily large scaling factor $\lambda^k$. 
Your concern about $\mu(B)/\mu(2B)$ being small compared to $\alpha$ can be  addressed by replacing $2$ with a  smaller scaling factor. 
Lemma. If $\mu$ is doubling, then for every $\epsilon>0$ there is $\lambda >1 $ such that for every ball $B$ we have 
$$\mu(\lambda B)< (1+\epsilon) \mu(B)$$
where as usual, $\lambda B$ means a scaled concentric ball.
To use the lemma, take $1+\epsilon = \alpha^{-1}$  and   conclude that $w(\lambda B)< \beta^{-1} w(B)$, which is the desired doubling property. So, we only need to prove the lemma. 
The proof, naturally, involves measuring some spherical shells. We can assume everything is centered at $0$. Write $A(r,R) = \{x: r< |x|\le R\}$. It turns out that adjacent spherical shells of equal thickness have about the same measure:  
Claim. If $\mu$ is doubling, then there exists   $M$ such that for every $r\ge 0$ and every $h>0$,
$$M^{-1}\le \frac{\mu(A(r,r+h))}{\mu(A(r+h,r+2h))} \le M\tag1$$
Indeed, let $\{p_i\}$ be a maximal $h$-separated subset of $A(r+h/3,r+2h/3)$. Being $h$-separated means $|p_i-p_j|\ge h$ for all $i\ne j$. Since the balls $B(p_i,h/3)$ are disjoint subsets of  $A(r,r+h)$, we have
$$\sum_{i} B(p_i,h/3) \le \mu(A(r,r+h))$$ 
The maximality of $\{p_i\}$ implies that the balls $  B(p_i, h )$ cover $A(r+h/3,r+2h/3)$. Hence, the balls 
$B(p_i, 3h )$ cover $A(r+h,r+2h)$. By the doubling condition, 
$$\mu(A(r+h,r+2h))\le \sum_{i} B(p_i, 3h ) \le  
C^4\sum_{i} B(p_i,  h/3 )\le C^4\mu(A(r,r+h))$$ 
This proves the first inequality in (1); the  other is proved in exactly the same way. $\qquad \Box$
The lemma is an easy consequence of the claim, so I'm going to wave hands a bit. Given a ball, say $B=\{|x|\le r\}$, divide it into two spherical shells of equal thickness, $A(0,r/2)$ and 
$A(r/2,r)$. According to the claim, each gets a "fair" share of the measure, hence the measure of each is appreciably smaller than the measure of the ball. Continuing this  division process, we find that $\mu(A(r-r/2^k,r))<\epsilon M^{-1}\mu(B)$ when $k$ is sufficiently large (it depends only on $M,\epsilon$). Using the claim again, obtain   $\mu(A(r,r+r/2^k))<\epsilon\mu(B)$. Thus, $\lambda = 1+2^{-k}$ works.
