Measuring subsets of N

Recently I tried to put a measure on the powerset of the natural numbers \mathcal{P}(\mathbb{N}), but it was pointed out in a comment that the map I defined is not even finitely additive.

Question: Is there a map \mu:\mathcal{P}(\mathbb{N}) \to [0, 1] with the following properties:

  • \mu(\mathbb{N}) = 1;
  • if A, B\subseteq\mathbb{N} such that A\cap B = \emptyset then \mu(A\cup B) = \mu(A)+\mu(B);
  • \mu is translation-invariant, that is \mu(A+n) = \mu(A) for A\subseteq \mathbb{N} and n\in\mathbb{N} where A+n = \{a+n: a \in A\}.

About dominiczypen

I'm interested in general topology, order theory, and graph theory. This link takes you to my preprints on arXiv.
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3 Responses to Measuring subsets of N

  1. Carlos says:

    The solution that occurs to me relies on the concept of Banach limit
    So choose any Banach limit $\Phi$, i.e., a bounded linear functional in the space of bounded sequences such that it is invariant under shifts. Then one just substitute the limit operator in the previous entry of the blog with the operator $\Phi$. You get immediately the three properties for $\mu$.

  2. vznvzn says:

    see from your recent post you are interested in P vs NP. lots of musings on that on my blog & also consider dropping by this chat room for further discussion

  3. Salvo says:

    Yes, and you can even enforce $\mu$ to be $(-1)$-homogeneous: see or Remark 3 in

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