Measure supported on finite subsets of $\mathbb N$ that is biased towards large numbers

Question

Let $\Omega$ be the family of all finite subsets of $\mathbb N$, and let $A_n\subset \Omega$ denote those sets which contain $n$. Does there exist a probability measure $\mu$ on $\Omega$ such that $\mu(A_n)\geq 1/2$ for all $n$?

Answer 1

Since $\Omega$ is countable, a probability measure is determined by just a list of numbers $\mu(\{x\})$ for each $x\in\Omega$ such that $\sum_{x\in\Omega}\mu(\{x\})=1$. But that sum means that for any $\epsilon>0$, there are finitely many $x_1,\dots,x_m\in\Omega$ such that $\mu(\{x_1\})+\dots+\mu(\{x_m\})>1-\epsilon$. Taking $\epsilon=1/2$, this means you have finitely many elements of $\Omega$ which together have over half of the total measure. Since each of these elements is a finite set, there exists $n\in\mathbb{N}$ which is not in $x_i$ for any $i\leq m$ (in fact, all but finitely many $n$ have this property), so $x_i\not\in A_n$ for each $i\leq m$. This means $\mu(A_n)\leq \mu(\Omega\setminus\{x_1,\dots,x_m\})<1/2$.

(More generally, this argument shows that for any $\epsilon>0$, $\mu(A_n)<\epsilon$ for all but finitely many $n$.)