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Let $0 \leq \alpha < \beta \leq 1$. I'm looking for an example of a Lebesgue measurable subset $E$ of $\mathbb{R}$ such that

$$\liminf_{\delta \rightarrow 0} \frac{m(E \cap (-\delta,\delta))}{2\delta} = \alpha$$


$$\limsup_{\delta \rightarrow 0} \frac{m(E \cap (-\delta,\delta))}{2\delta} = \beta$$

where $m$ is the Lebesgue measure on $\mathbb{R}$.

Can someone give an example? Thank you, Malik

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What about a measure on $[-1,1]$ wich can be described by a density as follows: $d(x)= 2$ if $x \in [1/(2n+1), 1/(2n)] \cap [-1/(2n), -1/(2n+1)]$ and $d(x)= 1/2$ if $x \in [1/(2n), 1/(2n-1)] \cap [-1/(2n-1), -1/(2n)]$ ? (density being the Radon–Nikodym derivative of the measure with respect to the Lebesgue measure) – Raskolnikov Dec 20 '10 at 13:17
Sorry, I was not paying attention. You were searching a set, not a measure. Anyway, Yuval gave a hint. – Raskolnikov Dec 20 '10 at 13:22
We already had a question like that, but I can't find it. – Yuval Filmus Mar 23 '11 at 0:27
(I deleted 2 comments that were posted on the merged question that no longer made sense. The comment of Yuval Filmus that presently precedes this one was originally posted on the merged question.…) – Jonas Meyer Mar 23 '11 at 0:57
up vote 4 down vote accepted

Try the duplication across zero of $$\bigcup_{n \geq 1} \left[\frac{1}{(2n)!}-\alpha\left(\frac{1}{(2n)!} - \frac{1}{(2n+1)!}\right),\frac{1}{(2n)!}\right] \cup \left[\frac{1}{(2n+1)!}-\beta\left(\frac{1}{(2n+1)!} - \frac{1}{(2n+2)!}\right),\frac{1}{(2n+1)!}\right].$$

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Thank you, nice solution! – Kalim Dec 24 '10 at 17:28

Divide the interval $(0,1]$ of radii into intervals which decrease "fast enough". In some of the intervals put some set of density $\alpha$, and in other put some set of density $\beta$. You can fill-in the rest of the details yourself.

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I think the following construction works: Given $\alpha \leq \beta$, define sequences $(a_n)_n$ and $(b_n)_n$ as follows: $$a_0 = 0;$$ $$b_0 = 1;$$ $$a_n = (\beta/\alpha)b_{n-1};$$ $$b_n = {1-\alpha \over 1-\beta}a_n.$$ Now define $E_n = [a_n, b_n)$ and $E = \bigcup_{n=0}^\infty E_n$.

The sequences were chosen so that $\bigcup_{i=0}^n E_i$ contains (roughly) $\beta$ of $[0, b_n)$ but only $\alpha$ of $[0, a_{n+1})$. It actually always contains slightly more than this, because it contains all of $[0, 1)$ instead of some crazy fractal pattern inside it, but any finite initial segment doesn't matter to the problem. As $R \rightarrow \infty$, the density of $E \cap [0, R)$ in $[0, R)$ oscillates (linearly!) between the two values, giving the behavior you requested.

But don't take this as gospel. It's been a while since I did any measure theory and I wasn't much good at it even at the time.

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