Here is an example. Let $d(A)$ be as you define it in the question, namely $$d(A)=\lim_{n\to \infty} \frac{|A\cap [1,n]|}{n}.$$ Let $U_n=\{k\in\mathbb{N} \text{ such that } k \text{ is a mutiple of } n!\}$. Then define $$\mu(A)=\lim_{n\to \infty} n!\cdot d(A\cap U_n).$$ Then I claim that $\mu$ is finitely additive and multiplicatively invariant. Finite additivity is obvious. For multiplicative invariance, note that for $s>k$, we have $(s+k)!\cdot d(kA\cap U_{s+k})=s!\cdot d(A\cap U_{s})$, unless I've screwed something up.
EDIT: Note by the way that one can replace the limit in the definition with $d$ with the Cesaro mean, for example, giving a much broader class of sets with defined measure. For example, with this addition, the set of natural numbers with a fixed leading digit in a fixed prime base $p$ has density $1/p$. 1/(p-1)$. 2 added 312 characters in body Here is an example. Let$d(A)$be as you define it in the question, namely $$d(A)=\lim_{n\to \infty} \frac{|A\cap [1,n]|}{n}.$$ Let $U_n=\{k\in\mathbb{N} \text{ such that } k \text{ is a mutiple of } n!\}$. Then define $$\mu(A)=\lim_{n\to \infty} n!\cdot d(A\cap U_n).$$ Then I claim that$\mu$is finitely additive and multiplicatively invariant. Finite additivity is obvious. For multiplicative invariance, note that for$s>k$, we have$(s+k)!\cdot d(kA\cap U_{s+k})=s!\cdot d(A\cap U_{s})$, unless I've screwed something up. EDIT: Note by the way that one can replace the limit in the definition with$d$with the Cesaro mean, for example, giving a much broader class of sets with defined measure. For example, with this addition, the set of natural numbers with a fixed leading digit in a fixed prime base$p$has density$1/p$. 1 Here is an example. Let$d(A)$be as you define it in the question, namely $$d(A)=\lim_{n\to \infty} \frac{|A\cap [1,n]|}{n}.$$ Let $U_n=\{k\in\mathbb{N} \text{ such that } k \text{ is a mutiple of } n!\}$. Then define $$\mu(A)=\lim_{n\to \infty} n!\cdot d(A\cap U_n).$$ Then I claim that$\mu$is finitely additive and multiplicatively invariant. Finite additivity is obvious. For multiplicative invariance, note that for$s>k$, we have$(s+k)!\cdot d(kA\cap U_{s+k})=s!\cdot d(A\cap U_{s})\$, unless I've screwed something up.