# Do the subsets of $\mathbb N$ that have asymptotic density form an algebra?

Consider $\Omega=\mathbb{N}.$ Is said that a $E\subset\mathbb{N}$ has a density limit if the following limit exists:

$$\rho(E)=\displaystyle \lim_{n\rightarrow \infty} \dfrac{\#\{k\in E: k\leq n\}}{n}$$

Let $\mathbb{F}$ the class of sets that have density. Is $\mathbb{F}$ an algebra?

## 1 Answer

No, because $\mathbb F$ is not closed under even finite intersections. Hint: Carefully construct a strictly and quickly increasing sequence $\{x_k\}_{k\in\mathbb N}$ of natural numbers and define $$C_k\equiv\{m\in\mathbb N\,|\,x_k<m\leq x_{k+1}\}$$ for each $k\in\mathbb N$. Let $A$ be the set of positive even integers in $C_1\cup C_3\cup\ldots$ together with the set of positive odd integers in $C_2\cup C_4\cup\ldots$. Moreover, let $B$ be the set of positive even integers.

Exercise: Show that $\{x_k\}_{k\in\mathbb N}$ can be constructed in such a way that $A,B\in\mathbb F$ but $A\cap B\notin\mathbb F$.

Source: Billingsley's Probability and Measure, Exercise 2.18.

• Consider an example sequence $x_k = 1000, 3000, 7000, 15000, \cdots,$. This would cause $\rho_n(A \cap B)$ to oscillate so that it never converges (where $\rho_n(E)$ is the $n^{th}$ term of the sequence under the limit). – Sudheer Jun 9 '15 at 12:29