Problem: If $\mu$ is a $\sigma$-finite measure on $(R,\mathscr{B}(R))$, then define $\mathscr{A}$ to be the collection of all $A\in\mathscr{B}(R)$ such that the following limit exists and is finite:$(D\mu)(A)=\lim_{n\rightarrow\infty}\frac{\mu(A\cap[-n,n])}{n}$. My question is: 1. is $\mathscr{A}$ an algebra? 2. If $R\in \mathscr{A}$, is $\mathscr{A}$ an algebra? 3. Is $D\mu$ countably additive on $(R,\mathscr{A})$? I just do not how to start it. $\mathscr{B}(R)$ means the Borel $\sigma$-algebra of $R$.
A problem about the limit of measure
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$\mathscr{A}$ is not even an algebra. Here is the counterexample. Let $A$ denote the set of even numbers and $B$ the set of even numbers in some $[4^n,2\cdot 4^n)$ and of odd numbers in some $[2\cdot 4^n,4^{n+1})$, for every $n$. Then, $A$ and $B$ both have density $\frac{1}{2}$ but the limit of $\frac{1}{k}\#(A\cap B\cap [0,k])$ does not exist. I will prove $(D\mu)(A)=\frac{1}{2}$. One has $(D\mu)(A)=lim_{n\rightarrow\infty}\frac{\mu(A\cap [-n,n])}{n}$. One has $\#(A\cap [-k,k])=\#(A\cap [0,k])=\lfloor{\frac{k}{2}}\rfloor+1$. Then, one has $(D\mu)(A)=lim_{k\rightarrow\infty}\frac{\lfloor{\frac{k}{2}}\rfloor+1}{k}=\frac{1}{2}$. BUT I have difficulty to prove $(D\mu)(B)=\frac{1}{2}$ AND $(D\mu)(A\cap B)$ OSCILLATE BETWEEN ROUGHLY $\frac{1}{6}$ AND ROUGHLY $\frac{1}{3}$ since I cannot provide the explicit formula like $(D\mu)(A)$.
I don't think $\mathcal{A}$ is necessarily a sigma-algebra, especially when $\mu$ is not finite, just sigma-finite. Here's an example. Take the counting measure on the positive integers. Then sets such as $P$, the set of primes has $(D\mu)(A)=0$, since by the prime number theorem there are roughly $n/\log n$ primes less than $n$. On the other hand consider the set $C$, the set of integers which do not have a $1$ as their first digit, e.g. $9,981$, are ok but $10$ is not. This set doesn't have a limit, as you can easily show (it does have a limsup). Since $C$ is the countable union of singletons, all of which are in $\mathcal{A}$, and limit 0, this means $\mathcal{A}$ is not a sigma algebra.