~~Strauch & Tóth [1]~~ Georges Grekos [3][4] showed that for any choice of upper and lower density, there is some subset of $\mathbb{N}$ with the chosen densities, provided the lower is no more than the upper and both are in [0, 1].


Mišík [2] extended this to show that for any choice of upper density, lower density, upper logarithmic density, and lower logarithmic density there is a subset of $\mathbb{N}$ with the chosen densities, provided they follow the necessary inequalities $0\leqslant\underline d\leqslant\underline\delta\leqslant\overline\delta\leqslant\overline d\leqslant 1.$


Is there a similar result with the uniform densities?
$$
\underline{u}(A)=\lim\_s\frac1s\liminf\_n\sum\_\stackrel{a\in A}{\ n < a\leqslant n+s}1
$$
$$
\overline{u}(A)=\lim\_s\frac1s\limsup\_n\sum\_\stackrel{a\in A}{\ n < a\leqslant n+s}1
$$


An ideal result would combine all three density types with the inequality
$$
0\leqslant\underline{u}\leqslant\underline{d}\leqslant\underline{\delta}\leqslant\overline{\delta}\leqslant\overline{d}\leqslant\overline{u}\leqslant1
$$
but I'm looking for any published results on the topic.


[1] Strauch and Tóth, Asymptotic density of $A\subset N$ and density of the ratio set $R(A)$. *Acta Arith.* **87** (1998), pp. 67-78. [eudml](https://eudml.org/doc/207205)


[2] Ladislav Mišík, Sets of positive integers with prescribed values of densities, *Mathematica Slovaca* **52**:3 (2002), pp. 289-296.[dml.cz](http://dml.cz/dmlcz/136863)


[3] Georges Grekos, Sur la répartition des densités des sous-suites d'une suite donnée, PhD thesis, Université Pierre et Marie Curie (1976).


[4] Georges Grekos, Répartition des densites des sous-suites d'une suite d'entiers, *J. Number Theory* **10**:2 (1978), pp. 177-191 (in French). [doi:10.1016/0022-314X(78)90034-3](http://dx.doi.org/10.1016/0022-314X(78)90034-3)



Prescribed values for the uniform density
OK. I think you can do this pretty easily by hand. 
First you need a way to generate sequences with very uniform density. The Sturmian sequences are perfect for this. A [Sturmian sequence](http://en.wikipedia.org/wiki/Sturmian_word) with parameter $\alpha$ has the property that sub-blocks of length $N$ have density converging to $\alpha$ uniformly in $N$.


Now: start by interspersing a Sturmian with parameter $\underline\delta$ with one of parameter $\bar\delta$. How to do the interspersing? Have one from $2^{n!}$ to $2^{(n+1)!}$. Then switch to the other between $2^{(n+1)!}$ and $2^{(n+2)!}$ etc. This switching is slow enough to guarantee that the sequence that you obtain has the prescribed $\underline\delta$ and $\bar\delta$. These are also the upper and lower densities for the time being.


Next we'll modify the sequence to obtain densities $\underline d$ and $\bar d$.
Alternately splice in segments of Sturmian parameter $\underline d$ and $\bar d$ between $2^{4^i}$ and $2^{4^i+2^i}$. This won't affect the upper and lower logarithmic densities (because $2^i/4^i\to 0$). 


But looking at these segments, you obtain a sequence with upper and lower densities $\underline d$ and $\bar d$ (the upper and lower uniform densities are also $\underline d$ and $\bar d$). 


Finally, we'll perturb things as in my comment to get the uniform densities we want.
For the segments between $2^n$ and $2^n+n$, insert alternately segments of the Sturmian sequences with densities
$\underline u$ and $\bar u$. These segments are so sparse, they will have no effect on the upper and lower densities or the logarithmic densities. They are enough to guarantee that you get the uniform densities you want.


And [Robert's your mother's brother](http://en.wikipedia.org/wiki/Bob%27s_your_uncle).


