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For $k\ge1$, let $A_k=\{0,k,2k,\ldots\}.$ Show that there is no measure $\mu$ on $\mathbb{N}$ satisfying $\mu(A_k)=\frac{1}{k}$ for all $k\ge1$.

What I have done so far:
I am trying to apply Borel-Cantelli lemma ($\mu(\mathbb{N})=1)$. Let $(p_n)_{n\in \mathbb{N}}=(2,3,5,\ldots)$ be the increasing sequence of all prime numbers. It is the case that for all $k\in\mathbb{N}$ and $i_1\lt i_2 \lt \ldots \lt i_k$ we have $\mu\left(A_{p_{i_1}}\cap\ldots\cap A_{p_{i_k}}\right)=\mu\left(A_{p_{i_1}}\right)\ldots\mu\left(A_{p_{i_k}}\right)$ so all $A_n$ are independent. We know that the sum of the reciprocals of the primes $\sum\limits_{n=1}^\infty \frac{1}{p_n}$diverges and hence, by B-C second lemma, $$\mu\left(\bigcap\limits_{n=1}^\infty\bigcup\limits_{m=n}^\infty A_{p_m}\right)=1$$ holds. However, I cannot figure out how to conclude the reasoning. I would appreciate any help.

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1 Answer 1

up vote 3 down vote accepted

You are basically there. You proved in your last line that $\mu$-a.e. number is divisible by infinitely many primes, which is an obvious contradiction.

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Oh, ok. I am almost sure I understand. Is the following $\mathbb{N}\subset \bigcup\limits_{n=1}^\infty\bigcap\limits_{m=n}^\infty A_{p_m}^c$ the precise form of what you have just stated? Also it is an easy contradiction, because the whole space cannot be contained in the set of measure $0$. Am I correct? –  Kuba Helsztyński Oct 29 '12 at 0:52
    
Yes, this is correct. –  Lukas Geyer Oct 29 '12 at 1:55

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