μ is sigma-finite measure on the Borel sets B(R) .

Question

$\mu$ is sigma-finite measure on the Borel sets $\bigl(\Bbb R,\mathcal{B}(\Bbb R)\bigr)$ , define $A$ is any collection of all $A$ in $\mathcal{B}(\Bbb R)$ such that the following limit exists and is finite, $$(D\mu)(A) = \lim\limits_{n\to\infty}\frac{\mu\bigl(A\cap[-n, n]\bigr)}{n}.$$ is A an algebra?

Answer 1

I believe the answer is "no" even when $\mu$ is the counting measure on $\mathbb N$.

That is: there are two sets $A, B \subset \mathbb N$ such that natural density exists for $A$ and for $B$, but not for $A \cap B$. That is: $$ \lim_n\frac{|A \cap \{1,2,\dots,n\}|}{n}\qquad\text{and}\qquad \lim_n\frac{|B \cap \{1,2,\dots,n\}|}{n} $$ both exist, but $$ \lim_n\frac{|A \cap B \cap \{1,2,\dots,n\}|}{n} $$ does not exist.

added

An example goes like this. Let $A$ be the even numbers. Choose a sequence $r_m$ increasing rapidly, like $4^m$ or $m!$. And for $B$ take

even numbers between $r_1$ and $r_2$,
odd numbers between $r_2$ and $r_3$,
even numbers between $r_3$ and $r_4$,
odd numbers between $r_4$ and $r_5$,
even numbers between $r_5$ and $r_6$,
odd numbers between $r_6$ and $r_7$,
and so on.

added
OK let's take the example with $r_m = 4^m$. Write $$ \lambda_n(E) = \frac{|E \cap \{1,2,\dots,n\}|}{n} $$ for any set $E$ and $n = 1, 2, 3, \dots$.

First note that $\lambda_{2n}(A) = \frac{n}{2n} = \frac{1}{2}$ and $\lambda_{2n+1}(A) = \frac{n}{2n+1}$ so in both cases $|\lambda_n(A) - \frac{1}{2}| < \frac{1}{n}$, and thus $\lambda_n(A) \to \frac{1}{2}$. Similarly, $\lambda_{2n}(B) = \frac{n}{2n}$ and $\lambda_{2n+1}(B)$ is either $\frac{n}{2n+1}$ or $\frac{n+1}{2n+1}$. So still for all $n$, we have $|\lambda_n(B) - \frac{1}{2}|<\frac{1}{n}$, and $\lambda_n(B)$ converges to $\frac{1}{2}$.

Now what about the intersection $A \cap B$? We will estimate $\lambda_{4^m}(A \cap B)$ taking the cases $m$ even and $m$ odd.

For $m$ even, we have $A \cap (4^{m-1},4^m] = B \cap (4^{m-1},4^m]$ is the set of all even numbers in $(4^{m-1},4^m]$. There are $\frac{3}{2}\cdot 4^{m-1}$ even numbers in that inverval. So $$ \lambda_{3^m}(A \cap B) \ge \frac{ \frac{3}{2}\cdot 4^{m-1}}{4^m} = \frac{3}{8} . $$

Now consider $m$ odd. Then $A \cap B \cap (4^{m-1},4^m]$ is empty, since no number is both even and odd. Thus $$ \lambda_{4^m}(A \cap B) \le \frac{|A \cap \{1,2,\dots,4^{m-1}\}|}{4^m} = \frac{\frac{1}{2} 4^{m-1}}{4^m} = \frac{1}{8} . $$

The sequence $\lambda_n(A \cap B)$ has a subsequence ${}\ge \frac{3}{8}$ and another subsequence ${} \le \frac{1}{8}$. So $\lambda_n(A\cap B)$ does not converge.