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If $I$ is a bounded interval in $\mathbb{R}$, let $2I$ denote an interval with the same center point but double the length.

A doubling measure on $\mathbb{R}$ is a (non-trivial, locally finite, Borel) measure $\mu$ such that $\mu(2I) \leq C\mu(I)$ for all intervals $I$ and some fixed constant $C$.

A doubling weight is an $L^1_{\text{loc}}$ function $w:\mathbb{R}\rightarrow\mathbb{R}_{\geq 0}$ such that $w dx$ is a doubling measure on $\mathbb{R}$.

Question: Is there a continuous doubling weight on $\mathbb{R}$ that vanishes on a set of positive Lebesgue measure?

Remarks:

  • There are many singular doubling measures on $\mathbb{R}$, not arising from doubling weights.
  • Without the assumption of continuity, there are doubling weights that vanish on sets of positive Lebesgue measure. In Section 1.8.8 of Stein, Harmonic Analysis an example (attributed to Journe) is given of a disjoint partition of $\mathbb{R}$ into sets $E_1$ and $E_2$ of positive measure such that the indicator functions $\chi_{E_1}$ and $\chi_{E_2}$ are both doubling weights. But this does not answer my question.
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  • $\begingroup$ Nate Eldredge: Thanks for the comment. Your measure is not doubling, but I think you were confused because I did not explain what I meant by $2I$ for an interval $I$. I have added an explanation. Now it is easy to see your measure is not doubling, and in fact that a doubling measure can never assign $0$ weight to an open interval. $\endgroup$
    – user100120
    Commented Oct 22, 2016 at 20:46
  • $\begingroup$ Oh, right. Silly of me. $\endgroup$
    – Nate Eldredge
    Commented Oct 22, 2016 at 20:46
  • $\begingroup$ I guess the obvious first candidate to try would be $w(x) = d(x,C)$ where $C$ is a fat Cantor set. Do you know whether that is doubling? $\endgroup$
    – Nate Eldredge
    Commented Oct 23, 2016 at 6:02
  • $\begingroup$ Nate Eldredge: This is a good suggestion, and modifications of it (e.g. raising to a power) work when $C$ is porous (hence measure zero). I thought about it a bit in this setting but couldn't quite see that it works. But maybe it does; I'll think about it further. I admit that mostly I'm surprised that, with doubling measures/weights studied for so long, that such a simple question would not be answered in the literature. $\endgroup$
    – user100120
    Commented Oct 23, 2016 at 12:52
  • $\begingroup$ @NateEldredge: There has been a study of such weights coming from metrics and indeed they are doubling.@user100120: I'm also amused with the luck of study on "continuous" doubling measure, although there is an extensive literature on the so-called $A_p$-weights and absolute continuity w.r.t. Lebesgue measure (these weights are automatically doubling). The connection is via something called $p$-Poincare inequalities and reverse Holder. My suggestion is that you like to look into these Muckenhapt $A_p$-weights to see of they are continuous. $\endgroup$
    – T. Amdeberhan
    Commented Oct 23, 2016 at 21:24

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