# Lebesgue measurable subset of $\mathbb{R}$ with given metric density at zero

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

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

but

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

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. math.stackexchange.com/questions/28577/…) – Jonas Meyer Mar 23 '11 at 0:57

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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