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I'm interested in the following property, for a positive and locally bounded function $\omega:\mathbb{R}\to\mathbb{R}^d$, $d\ge 1$: there exists a countable sequence of open and pairwise disjoint sets $A_j$, such that

  1. $\mathbb{R}^d\setminus\bigcup_{j=1}^\infty A_j=N$, with $\mathcal {L} ^d(N)=0$ ($\mathcal L^d$ is the $d$-dimensional Lebesgue measure),
  2. $a_j\le\omega(x)\le b_j$ , for any $x\in A_j$, with $\frac{b_j}{a_j} \le C$ for some constant $C>0$ and for any $j$.

Since it is pretty involved and abstract, as a definition, I would like to have some criterion that ensures it to be true. At the moment I've understood the following:

  1. if $C_1\le\omega(\cdot)\le C_2$, with $C_1,C_2 >0$ then the property is obviously satisfied, with $A_1=\mathbb R^d$;
  2. taking $A_j=\omega^{-1}\big(]2^n,2^{n+1}[\big)$ , $n\in \mathbb Z$, the property is true, with $C=2$, for radial functions with continuous and injective radial part (since $N=\bigcup_n \omega^{-1}(2^n)$ is a union of spheres of dimension $d-1$);
  3. since if $\mathcal L^d(\omega^{-1}(2^n))>0$ then $2^n$ has to be a critical value, by the Sard Lemma, if we move a little the extrema of the above intervals, the property is true also for $\mathcal C^d(\mathbb R^d)$ functions ;
  4. if $f(\cdot)\le\omega(\cdot)\le g(\cdot)$, with $f$ and $g$ satisfying this property, then also $\omega$ satisfies this property.

Due to the very bad behavior of continuous functions, I think that I cannot extend directly the arguments of point 2 and 3 to general continuous functions (which is where I hoped this would be true). Is this true?

If this is the case, then there are some results that says when can I bound $\omega$ with $\mathcal C^d(\mathbb R^d)$ or continuous and radially injective functions?

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Erm... Isn't it just a fancy way to say that $\log\frac{\limsup_x\omega}{\liminf_x\omega}\in L^\infty$? –  fedja Sep 10 '12 at 23:15
    
It can be, but I do not completely get it. This property implies immediately that $\frac{\limsup_x \omega}{\liminf_x \omega}\in L^\infty$. Reasoning on a compact covering of $\mathbb R^d$, it seems to me that this is indeed a characterization. Why do you say that the $\log$ is needed? Sorry but my measure theory is pretty rusty! –  Dario Prandi Sep 11 '12 at 10:10
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Yes, it seems that one characterization is $\frac{\lim_{r \to 0}\sup_{B(x,r)} \omega}{\lim_{r \to 0}\inf_{B(x,r)} \omega}\in L^\infty$. (The converse follows by standard covering arguments.) Log does not make any difference, since this is always bounded below by $1$. –  Tapio Rajala Sep 11 '12 at 10:56
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1 Answer

In the case $\omega \colon \mathbb R^d \to \mathbb R$ is continuous, you can take $A_j = \omega^{-1}(]t_j,t_{j+1}[)$ with $t_i \in ]2^i,2^{i+1}[$ chosen so that $\mathcal L^d(\omega^{-1}(t_i)) = 0$ for all $i \in \mathbb Z$. The existence of such $t_i$ follows since $(\omega^{-1}(t))_{t \in \mathbb R}$ are disjointed and only for countably many $t \in \mathbb R$ they can have positive $\mathcal L^d$-measure.

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