Prove that there is no probability measure on $\mathbb{N}$ Prove that there is no probability measure on $\mathbb{N}$ such that for any $n\geq 1$, the probability of the set of multiples of $n$ is $1/n$.

I'm trying to proof this by contradiction, but actually I have no clue so far. Much appreciated if someone can help.
Suppose there were such a measure, $\mu$. $\newcommand\of[1]{\left({#1}\right)}\newcommand\N{\Bbb{N}}$

Let $X_n=\{m\in\Bbb{N} : n\mid m\}$ be the set of multiples of $n$. Certainly these sets must be measurable. 

First let's deal with a simple case, computing 
$\mu(X_{p_1}\cup \cdots \cup X_{p_i})$, where the $p_i$ are all distinct primes. Since the primes are all distinct, we get that finite intersections of sets have measure the product of the measures of the individual sets.
By PIE, we then get that 
$$
\mu(X_{p_1}\cup\cdots \cup X_{p_i}) 
= \sum_{j=1}^i \frac{1}{p_j} - \sum_{j_1&lt;j_2} \frac{1}{p_{j_1}p_{j_2}} + \cdots + (-1)^{i+1} \frac{1}{p_1p_2\cdots p_i}.
$$
Now, if we subtract both sides from 1, then the right hand side factors to give
$$
1-\mu(X_{p_1}\cup\cdots \cup X_{p_i})
= \prod_{j=1}^i \of{1-\frac{1}{p_j}},
$$
so 
$$\mu(X_{p_1}\cup\cdots \cup X_{p_i}) = 1- \prod_{j=1}^i \of{1-\frac{1}{p_j}}
=1-\frac{\phi(p_1\cdots p_i)}{p_1\cdots p_i}.$$


  Now we make an observation: For any sequence $n_1,\ldots,n_k,\ldots$ containing all but finitely many prime numbers, 
  $$\mu\of{\bigcup_{k\in\N} X_{n_k}} = 1.$$


Proof.

We can assume that the $n_i$ are all prime, since adding extra sets only increases the measure. The result then follows from the previous result combined with the following fact:
$$\prod_{p\text{ prime}} \of{1-\frac{1}{p}} = 0.$$

To prove this fact, take $-\log$ of a partial product across primes below some upper bound, $p&lt;N$, which gives 
$$\sum_{p &lt; N} -\log(1-1/p) = \sum_{p&lt;N}\sum_{n=1}^\infty \frac{1}{p^nn} \ge \sum_{p&lt;N} \frac{1}{p}.$$
The reciprocals of the primes are well known to be a divergent series. 
(Most proofs of this actually go the other direction and prove that this product diverges, but I'm just trying to get a quick reduction to a well known fact, since this is not the point of my answer).
Thus $-\log$ of the partial products go to $\infty$, which implies that the partial products converge to $0$.

This completes the proof of this fact. $\blacksquare$

Now we can show that if all the $X_n$ are measurable and there is a measure with value $\mu(X_n)=1/n$, then for all naturals $m$, $\{m\}$ is measurable and has measure $0$, which contradicts the claim.

Proof.

Induct on $m$ by divisibility. 
First $\{1\}$ is the complement of 
$$\bigcup_{p\text{ prime}}X_p,$$ which we just showed has measure $1$, so our base case holds.

For $m&gt;1$, assume the result for all divisors of $m$. Then 
$$\mu(\{ d : d\mid m\}) = \mu\of{\Bbb{N} \setminus \bigcup_{d\nmid m} X_d}=1-1=0.$$
By the inductive hypothesis, $\mu(\{d:d\mid m\}) = \mu(\{m\})$, so $\mu(\{m\})=0$, as claimed.

Finally, for $m=0$, since $\{0\}=\bigcap_{n\in\N} X_n$, we have $\{0\}$ is measurable and $\mu(\{0\}) &lt; 1/n$ for all $n$, which implies $\mu(\{0\})=0$. 

Thus the claim is proved. $\blacksquare$
I believe such a probability measure would need to be uniform, so we have two cases: either $P(X=n)=0$ for all $n \in \mathbb{N}$, or $P(X=n)=\epsilon&gt;0$ for all $n \in \mathbb{N}$.  

If $P(X=n)=0$, then $1=P(\mathbb{N})=\sum_{n=1}^{\infty} P(X=n)=0$, a contradiction.  

If $P(X=n)=\epsilon&gt;0$, then $1=P(\mathbb{N})=\sum_{n=1}^{\infty} P(X=n)=\sum_{n=1}^{\infty} \epsilon=+\infty$, another contradiction.  

Therefore, $\mathbb{N}$ cannot have a uniform probability measure.