A crazy idea...


Let $\mathcal{B}$ be the Borel subsets of $\mathbb{R}$, and $X$ the set of continuous functions $[0,\infty) \to \mathbb{R}$. Does there exist a "measure" function $\mu: \mathcal{B} \to X$ satisfying the following conditions?

  1. $\mu(\varnothing) = \boldsymbol{0}$.

  2. For all $d \in [0,\infty)$, if $A \in \mathcal{B}$ and $\mu_d$ is $d$-dimensional Hausdorff measure, then: $$\lim_{x \to \infty} \frac{\mu(A)(x)}{x^d} = \mu_d(A).$$ (Specifically if $A$ has Hausdorff dimension $d$ and measure $a$ in that dimension, we want it to grow like $ax^d$, if $a \ne 0, \infty$.)

  3. $\mu$ is countably additive, in the following sense: let $A_1, A_2, A_3, \ldots$ be disjoint Borel sets with union $A$, $\mu(A_n) = f_n$ and $\mu(A) = f$. If $\sum_{n=1}^\infty f_n$ converges uniformly on compact subsets of $\mathbb{R}$, then

$$ \sum_{n=1}^\infty f_n = f. $$

Update: Eric Wofsey has shown below by an elementary argument that if $\mu$ also satisfies:

  1. Translation-invariance: $\mu(A + r) = \mu(A)$ for all $A \in \mathcal{B}$ and $r \in \mathbb{R}$.

then this is impossible. What about 1-3?


While Lebesgue measure on $\mathbb{R}$ fails to capture any size differences between different measure-zero sets, we can capture such information with the Hausdorff dimension. By identifying a set with its Hausdorff dimension and its Hausdorff measure in that dimension, we get a much richer notion of size.

For example, $\varnothing$ (dimension 0 measure 0) is smaller than $\{1,4\}$ (dimension 0 measure $2$), which is smaller than $\mathbb{Q}$ (dimension 0 measure $\infty$), which is smaller than the Cantor set (dimension $\frac{\ln 2}{\ln 3}$ measure $1$) which is smaller than $[0,1]$ (dimension $1$ measure $1$), which is smaller than $\mathbb{R}$ (dimension $1$ measure $\infty$).

I have long been curious: can we capture this notion of size in a single "measure" on Borel sets -- not real-valued, but function-valued? Above is a specific attempt.

  • $\begingroup$ Is there a reason you haven't written axiom 2 as $\lim\frac{\mu(A)(x)}{x^d}=a$ and then allowed $a$ to be $0$ or $\infty$? In fact, with that version, you could just require $\lim\frac{\mu(A)(x)}{x^d}=\mu_d(A)$ for all $d$ (where $\mu_d$ is $d$-dimensional Hausdorff measure), without needing to mention the Hausdorff dimension of $A$ directly. $\endgroup$ – Eric Wofsey Aug 16 '16 at 8:28
  • $\begingroup$ @EricWofsey Thanks, that's a very good point. I didn't know exactly what to do with measure $0$ or $\infty$ sets so I left out any conditions for them, but your approach seems more natural. I've updated axiom 2 to be what you suggested. $\endgroup$ – 6005 Aug 16 '16 at 20:12
  • $\begingroup$ FYI, the middle-thirds Cantor set has measure $1$ in dimension $\frac{\ln 2}{\ln 3}.$ $\endgroup$ – Dave L. Renfro Aug 16 '16 at 20:41
  • $\begingroup$ @DaveL.Renfro Thanks -- I was curious. $\endgroup$ – 6005 Aug 16 '16 at 20:43
  • 1
    $\begingroup$ It took me a few minutes, but I finally found a stackexchange question/answer that shows this measure is $1$ -- How prove that $m_\alpha ^*(\mathcal C)\geq 1$ where $\mathcal C$ is the Cantor set, and $m_\alpha $ the Hausdorff measure. $\endgroup$ – Dave L. Renfro Aug 16 '16 at 20:52

This is impossible if you require translation-invariance. Indeed, consider $A=\mathbb{N}$, $B=\{0\}$, and $C=A\setminus B$. Then translation-invariance implies $\mu(A)=\mu(C)$, since $C=A+1$. But axioms (1) and (3) imply that $\mu(-)(x)$ is finitely additive for each $x$, so $\mu(A)(x)=\mu(B)(x)+\mu(C)(x)$ for all $x$. Thus $\mu(B)(x)=0$ for all $x$. This contradicts axiom (2), since $\mu(B)(x)$ must converge to $1$ as $x\to\infty$.

(Note that ordinary $0$-dimensional Hausdorff measure avoids this problem by having $\mu(A)$ and $\mu(C)$ be infinite, so you can't subtract and deduce that $\mu(B)=0$.)

  • $\begingroup$ Thank you. I will leave the question in case 1-3 are still satisfiable. $\endgroup$ – 6005 Aug 17 '16 at 3:37

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