The natural density of the set $A \subseteq \mathbb{N}$ is defined as $$\delta(A)=\lim_{n \rightarrow \infty}\frac{ \# \{k \in A | k \leq n \} }{ \# \{k \in \mathbb{N} | k \leq n \} }.$$ My idea was to generalize this to sets with greater cardinality:
For example take the set $B = \{x \in \mathbb{C} | \ \lvert x \rvert \leq \frac{1}{2} \}$ instead of $A$ and the unit disk $U = \{x \in \mathbb{C} | \ \lvert x \rvert \leq 1 \}$ instead of $\mathbb{N}$. Now we know by some easy geometrical calculations that the ratio is $\frac{1}{4}$. How do we get this by a concept similar to the natural density or other densities with countable sets?

Is there a density concept (with limits) for general sets, including uncountable ones like $\mathbb{R}$?

Thanks for your ideas, comments and answers!

  • $\begingroup$ You may use measures, but I am not sure that it will well-behaved and enjoy the properties you may want. $\endgroup$
    – Hanul Jeon
    Jan 7, 2021 at 16:57

1 Answer 1


A density always requires a superset of the considered set. Let's denote the superset of the considered set $A$ in the following by $B$. To transfer your concept to any other countable set should be easy, since for each countable set $B$ exists a injective function $f:B\to\mathbb N$. In this case just consider $f(A)$ resp. $f(B)$ instead of $A$ resp. $\mathbb N$ and proccede as you already did $$\delta_f(A):=\lim_{n \rightarrow \infty}\frac{ \# \{k \in f(A) | k \leq n \} }{ \# \{k \in f(B) | k \leq n \} }. $$ I think that it is obvious that it's important to take the same function $f$ (in the numerator and the denominator) for both sets $A,B$ although a lot of such functions $f$ exit. This density is not independent of the choice of $f$. I will give an example for this at the end of this answer.

To transfer your concept to an uncountable set one has to analyze what you have done so far: You use the counting measure when you were dealing whith natural numbers. This means you have to choose a new measure on the new superset $B$ you are now dealing with. As you mentioned $\mathbb C\cong\mathbb R^2$ the Lebesgue measure $\lambda$ would be a 'natrual' choice, but you can choose of course different. I think it is obvious that the density depends on the measure. What you were doing with the limit is, you intersected both sets with an sequence $\mathcal C:=(C_n)_{n\in\mathbb N}$ of increasing sets $C_n$ with the properties $$\forall n\in\mathbb N:C_n\subseteq C_{n+1}\tag{1}\\ B\subseteq\bigcup_{n\in\mathbb N}C_n. $$ These properties are just important if $\lambda(B)=\infty$, as measures satisfy a continuity property, you wouldn't need any intersection in the case of $\lambda(B)<\infty$. Otherwise you don't need this sequence. In your special case $A=\{x \in \mathbb{C} | \ \lvert x \rvert \leq \frac{1}{2} \}$ and $B= \{x \in \mathbb{C} | \ \lvert x \rvert \leq 1 \}$ you can directly use (with the Lebesgue measure) $$\delta_{B,\lambda}(A)=\frac{\lambda(A)}{\lambda(B)}=\frac{\pi/4}{\pi}=\frac{1}{4} .$$ In the general case this means $$\delta_{B,\lambda,\mathcal C}(A):=\lim_{n \rightarrow \infty}\frac{\lambda(A\cap C_n)}{\lambda(B\cap C_n)} .$$ In the space $\mathbb R^d$ you could use for example the balls with radius $n$ as $C_n=\{x\in\mathbb R^d: ||x||_2\leq n\}$. This density is in general not independent of the choice of the sequence $\mathcal C=(C_n)_{n\in\mathbb N}$ although measures satisfy a continuity property. I will give an example for this at the end of this answer.

Back to my first example above, this concept means you define a counting measure on the set $B$ by $\lambda(B):=\#f(B)$ (or directly $\lambda(B):=\# B$, as it is the counting measure where all elements have the same weight and $f$ is injektive, it makes no difference) and $\mathcal C_f=(C_n)_{n\in\mathbb N}$ with $C_n:=\{b\in B: f(b)\le n\}$: $$\delta_{B,\lambda,\mathcal C_{f}}(A):=\lim_{n \rightarrow \infty}\frac{\lambda(A\cap C_n)}{\lambda(B\cap C_n)}=\lim_{n \rightarrow \infty}\frac{ \# \{k \in f(A) | k \leq n \} }{ \# \{k \in f(B) | k \leq n \} }=\delta_f(A). $$ From this you see that sequence of the sets $C_n$ is affected by the choice of $f$. This is the reason why the density $\delta_f=\delta_{B,\lambda,\mathcal C_{f}}$ depends on $f$. To give a simple example consider the set $B=\mathbb N$ and the two different choices $f_1=\operatorname{Id}_{\mathbb N}$ and $f_2$, whereas $f_2$ takes always the lowest two odd numbers mapping to the next two natural numbers then take the next lowest even number mapping to the next natural number and so on, i.e. $$f_2:B\to\mathbb N:(1,3,\color{green}{2},5,7,\color{green}{4},9,11,\color{green}{6},\dots)\mapsto(1,2,\color{green}{3},4,5,\color{green}{6},7,8,\color{green}{9},\dots).$$ Now consider the set $A$ consisting of all even natural numbers. I think it's obvious to see that $$\delta_{f_1}(A)=\delta_{B,\lambda,\mathcal C_{f_1}}(A)=\frac{1}{2}\\\text{and }\delta_{f_2}(A)=\delta_{B,\lambda,\mathcal C_{f_2}}(A)=\frac{1}{3}.$$ So this was the example of which you see that the density $\delta_f$ of the beginnig is dependent on the function $f$ and the general density $\delta_{B,\lambda,\mathcal C}$ is depnendent on the sequence $\mathcal C$, which is not obvisous in the first view. The other dependencies $B,\lambda$ are easier to see.


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