I'm trying to solve the following exercise from Hjort's measure theory notes:
Let $X=2^{\mathbb{N}}$, the collection of all functions from $\mathbb{N}$ to $\{0,1\}$ in the product topology.
Show that for each $N$ and $r\in\mathbb{R}$, the collection $A_{N,r}$ of $f\in X$ with $\dfrac{1}{N}|\{n<N:f(n)=1\}|>r$ is Borel.
My problem is that i don't know how to approach it. Also I'm confused about the definition of $A_{N,r}$. Any help or hints are welcome.
By $A_{N,r}$ I think he means $\{f\in X: \frac{1}{N}|\{n<N:f(n)=1\}|>r\}$ which can be rewritten as $\{f\in X: |\{n<N:f(n)=1\}|>Nr\}$. As $r$ can be any number, the number $Nr$ could be anything so we might as well call it $K$. But, we can rewrite $\{f\in X: |\{n<N:f(n)=1\}|>K\}$ as $\bigcup\limits_{I\subset \{1, 2,\cdots,N\}: |I|>K}\{f\in X : f(n)=1 \text{ for }n\in I\}$. It should be pretty clear that these sets are measurable when $\{0,1\}^{\mathbb{N}}$ is equipped with the product $\sigma-$algebra of the discrete $\sigma-$algebra on $\{0, 1\}$ (or, what is equivalent, the Borel $\sigma-$algebra induced by the product topology on $\{0,1\}^{\mathbb{N}})$, since they consists of all $f$s that attain some specified value on a finite number of points. Moreover, we are only taking the union over finitely many sets, so the union is itself measurable.