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Dec 5, 2018 at 11:08 comment added YCor Why don't you correct, amend or delete your post, since it's been mentioned that it's wrong?
Dec 4, 2018 at 17:16 comment added Robert Furber For the record, $\mu$ is not finitely additive. We can build a subset $A$ of $\mathbb{N}$ such that the limsup is strictly larger than the liminf. Then, because $1-x$ is an order-reversing isomorphism of $[0,1]$, $$\mu(\mathbb{N}\setminus A) = 1 - \limsup_{n \to \infty} \frac{|A \cap \{1,\ldots,n\}|}{n} < 1 - \mu(A)$$.
Dec 3, 2018 at 9:36 comment added Dirk Your algebra $\mathcal{A}$ is $\mathcal{P}(\mathbb{N})$, right? But this is already a $\sigma$-algebra, so the extension of $\mu$ is $\mu$ itself which is not countable additive, or do I miss something?
Dec 3, 2018 at 9:24 history answered Dominic van der Zypen CC BY-SA 4.0