# Math Help - Limit of this counting measure

1. ## Limit of this counting measure

Let $A \subset \mathbb {N}$, define $\mu (A) = \lim _{n \rightarrow \infty } \frac { count [A \cap \{ 1,...,n \} ] } {n}$

where "count" denote the number of elements in that set.

Let M be the set of all even numbers, N be the set of all odd numbers, and P be the set of all perfect squares.

Q1: Find $\mu (M) \ , \ \mu(N) \ , \ \mu (P)$

Q2: If A and B are disjoint sets, show that $\mu (A \cup B) = \mu (A) + \mu (B)$

Q3: Is it a measure?

My solutions so far:

Q1: For $\mu (M) = \lim _{n \rightarrow \infty } \frac { count [M \cap \{ 1,...,n \} } {n}$, and I have $count [M \cap \{ 1,...,n \} = \frac {n}{2}$ since only half of those will be even, so the whole limit would equal to $\frac {1}{2}$, same goes for all odd numbers.

For the perfect squares, should it be $\frac { \sqrt {n} }{n} \rightarrow 0$?

And if I can prove Q2, doesn't that automatically prove Q3?

Thank you!

2. ## Re: Limit of this counting measure

Let $A \subset \mathbb {N}$, define $\mu (A) = \lim _{n \rightarrow \infty } \frac { count [A \cap \{ 1,...,n \} ] } {n}$

where "count" denote the number of elements in that set.

Let M be the set of all even numbers, N be the set of all odd numbers, and P be the set of all perfect squares.

Q1: Find $\mu (M) \ , \ \mu(N) \ , \ \mu (P)$

Q2: If A and B are disjoint sets, show that $\mu (A \cup B) = \mu (A) + \mu (B)$

Q3: Is it a measure?

My solutions so far:

Q1: For $\mu (M) = \lim _{n \rightarrow \infty } \frac { count [M \cap \{ 1,...,n \} } {n}$, and I have $count [M \cap \{ 1,...,n \} = \frac {n}{2}$ since only half of those will be even, so the whole limit would equal to $\frac {1}{2}$, same goes for all odd numbers.

For the perfect squares, should it be $\frac { \sqrt {n} }{n} \rightarrow 0$?

And if I can prove Q2, doesn't that automatically prove Q3?

Thank you!
Q1's answer looks good. And, you tell us, does it? Are they the only two axioms an (outer) measure satisfies? Moreover, I hope it's clear how to prove Q2.