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I was wondering if there exists a non trivial outer measure on the natural numbers $\mathbb N$.

$\mu(A) = 0$ for all $A\in\mathfrak P(\mathbb N)$

Is certainly monotonic, $\sigma$-sub additive and $\mu(\emptyset) =0$ and therefore an outer measure. But when I tried to construct an outer measure like

$\mu(A) = 0$ if $\#A<\infty$ and $\mu(A) = k$ if $\#A=\infty$ with $k\in \mathbb N \cup \infty$

I noticed, that these won't work because

$k = \mu(\mathbb N) = \mu\left(\bigcup_{n=1}^\infty [n, n+1]\right) \overset{\sigma}{\leqslant} \sum_{n=1}^\infty \mu([n, n+1]) = 0.$

Using $\mu([a,b]) = b-a$ won't work either because $\mathbb N$ is countable and $\bigcup_{n=1}^\infty [n,n] = \mathbb N$.

Could you give an example, way of construction or show non-existence?

Thank you

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    $\begingroup$ Take $\mu(A)=\sum_{n\in A}\frac{1}{2^{n+1}}$. $\endgroup$
    – user700480
    Sep 24, 2022 at 8:35
  • $\begingroup$ Thank you @StinkingBishop! $\endgroup$
    – HelloWorld
    Sep 24, 2022 at 9:07

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You can consider the application $\mu^*\colon \mathcal{P}(\mathbb{N})\to [0,+\infty]$ defined as $$\mu^*(E)=0\quad\text{if}\quad E=\emptyset$$ $$\mu^*(E)=1\quad\text{if}\quad E\ne\emptyset$$

This is an axample of an outer measure on $\mathbb{N}$. However, note that it is not a measure.

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  • $\begingroup$ Thank you @NatMath! $\endgroup$
    – HelloWorld
    Sep 24, 2022 at 9:08

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