# Outer measure on $\mathbb N$

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

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

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.