On the independence of lower and upper asymptotic and Banach densities Given a set $X \subseteq \mathbf N^+$, denote by $\mathsf{d}_\ast(X)$ and $\mathsf{d}^\ast(X)$, respectively, the lower and upper asymptotic (or natural) density of $X$, viz. $$\mathsf{d}_\ast(X) := \liminf_{n \to \infty} \frac{|X \cap [1, n]|}{n}$$ and $$ \mathsf{d}^\ast(X) := \limsup_{n \to \infty} \frac{|X \cap [1, n ]|}{n},$$ and by $\mathsf{bd}_\ast(X)$ and $\mathsf{bd}^\ast(X)$, respectively, the lower and upper Banach (or uniform) density of $X$, i.e. $$\mathsf{bd}_\ast(X) := \lim_{n \to \infty} \min_{m \ge 1} \frac{|X \cap [m+1, m+n ]|}{n}$$ and $$ \mathsf{bd}^\ast(X) := \lim_{n \to \infty} \max_{m \ge 1} \frac{|X \cap [m+1, m+n]|}{n}.$$ It is mentioned in @Martin Sleziak's answer to Question 66191: Density of a set of natural numbers whose differences are not bounded that for all $\alpha, \beta, \gamma, \delta \in [0,1]$ with $\alpha \le \beta \le \gamma \le \delta$ there exists a set $X \subseteq \mathbf N^+$ such that $\mathsf{bd}_\ast(X) = \alpha$, $\mathsf{d}_\ast(X) = \beta$, $\mathsf{d}^\ast(X) = \gamma$, and $\mathsf{bd}^\ast(X) = \delta$.

Q. Do you know of a reference for the claim above?

Notice that Martin is not sure about the claim (he writes, "If I remember correctly, I have seen the result etc."), and from the comments to his own answer it seems he could not indeed retrieve a reference.
I am aware of results along these lines in the case where the pair $(\mathsf{d}^\ast, \mathsf{bd}^\ast)$ and its "dual" $(\mathsf{bd}_\ast, \mathsf{d}_\ast)$ are replaced, respectively, with the pair $(\mathsf{ld}^\ast, \mathsf{d}^\ast)$ and its dual $(\mathsf{d}_\ast, \mathsf{ld}_\ast)$, where $\mathsf{ld}_\ast$ and $\mathsf{ld}^\ast$ are, respectively, the lower and upper logarithmic density on $\mathbf N^+$, namely $$\mathsf{ld}_\ast(X) := \liminf_{n \to \infty} \frac{\sum_{x \in X \cap [1,n]} \frac{1}{x}}{\log n}.$$ and $$\mathsf{ld}^\ast(X) := \limsup_{n \to \infty} \frac{\sum_{x \in X \cap [1,n]} \frac{1}{x}}{\log n}.$$ In fact, this can even be proven in the general case where the pair $(\mathsf{ld}^\ast, \mathsf{d}^\ast)$ is replaced by a pair $(\mu^\ast, \nu^\ast)$ of $\mathfrak{m}$-weighted upper densities on an arithmetical semigroup $\mathbb{A} = (A, \cdot\,, |\cdot|)$ and $(\mathsf{d}_\ast, \mathsf{ld}^\ast)$ by the dual pair of $(\mu^\ast, \nu^\ast)$, under the assumption that $\mu^\ast(X) \le \nu^*(X)$ for all $X \subseteq A$ and $$\sum_{a \in A,\, |a| \le x} \mathfrak{m}(a) \sim x^s$$ for some $s > 0$ as $x \to \infty$, see 

F. Luca, C. Pomerance, and Š. Porubský, Sets with prescribed arithmetic densities, Unif. Distr. Theory 3 (2008), No. 2, 67-80

for further details. Yet, I could not find a reference for the case I am asking for.
 A: So here is an explicit construction that works for any $\alpha\le\beta\le\gamma\le\delta$.
I first claim that if you can produce a set $Y$ such that bd$_*(Y)=d_*(Y)=\beta$, bd$^*(Y)=d^*(Y)=\gamma$, then it's easy to modify it to produce a set $X$ with the required properties.
First, notice that by unique ergodicity of the circle rotation $x\mapsto x+\pi\pmod 1$, for any $\rho\in [0,1]$, $S_\rho=\{n\colon \langle n\pi\rangle\le \rho\}$ has upper and lower Banach density equal to $\rho$ (where $\langle x\rangle$ is the fractional part of $x$). Now given $Y$, modify it on $[2^{2n},2^{2n}+2n]$ to match $S_\alpha$ and on $[2^{2n+1},2^{2n+1}+2n+1]$ to match $S_\delta$. 
That is:
$$
X=\left(Y\setminus\bigcup_n[2^n,2^n+n]\right)\cup\bigcup_{n\text{ even}}
(S_\alpha\cap[2^n,2^n+n])\cup
\bigcup_{n\text{ odd}}(S_\delta\cap[2^n,2^n+n]).
$$
Clearly $d_*(X)=d_*(Y)$ and $d^*(X)=d^*(Y)$ since the set has been modified on a vanishingly small proportion of the integers. Let $A=\mathbb N\setminus\bigcup_n[2^n,2^n+n]$.
By the unique ergodicity, it is clear that bd$_*(X)\le\alpha$ and bd$^*(X)\ge\delta$. On the other hand, given a long interval $J$, if $J$ intersects $A$ on a large part sub-interval, $K$, the density $X$ on $K$ is at least $\beta-\epsilon$, while on the other substantial parts the density is at least $\alpha-\epsilon$. This shows bd$_*(X)\ge\alpha$, so that bd$_*(X)=\alpha$ and similarly bd$^*(X)=\delta$.
Hence it suffices to construct a $Y$ with the correct upper and lower densities. Such a $Y$ is given by
$$
Y=\bigcup_{n\text{ even}}\left(S_\beta\cap [2^{2^n},2^{2^{n+1}})\right)
\cup
\bigcup_{n\text{ odd}}\left(S_\gamma\cap [2^{2^n},2^{2^{n+1}})\right).
$$
