Complexity of ultrafilter limits Let $\mathscr{F}$ be a free ultrafilter on $\mathbf{N}$ and, for each $A\subseteq \mathbf{N}$ and $n \in \mathbf{N}$, define
$$
d_n(A):=\frac{|A\cap [1,n]|}{n}.
$$

Question. Considering $\mathcal{P}(\mathbf{N})=\{0,1\}^{\mathbf{N}}$, is it true that the set
  $$
\{A\subseteq \mathbf{N}: \mathscr{F}\text{-}\lim d_n(A)=0\}
$$
  is not Borel?

 A: Yes. Let me rather write subsets as sequences: the question is whether the subset $L$ of $(a_n)$ in $\{0,1\}^{\mathbf{N}}$ such that $\mathscr{F}\text{-}\lim\sum_{k=1}^n\frac{a_k}{n}=0$ is non-Borel.
Write $u_n=2^{2^n}$ (essentially all that matters is $u$ increases and that $\sum_{k<n}u_k=o(u_n)$). Write $\Phi(a)_n=(a_{u_n})$. Let $P\subset\{0,1\}^{\mathbf{N}}$ be the subset of sequences that are constant on $[u_n+1,u_{n+1}]$ for all $n$. Then $\Phi$ induces a homeomorphism from $P$ to $\{0,1\}^{\mathbf{N}}$. Let $v$ be defined by $v(0)=0$, $v(u_n+1)=v(u_n)+1$, and $v(k+1)=v(k)$ is $k$ is not among the $u_n$ (thus $P$ is the set of sequences that are constant on fibers of $v$). Let $\omega$ be the image of $\mathscr{F}$ by $v$. Then $\Phi(L\cap P)$ is simply the set of sequences whose $\omega$-limit is zero. When $\{0,1\}$ is meant modulo 2, this is a dense subgroup of index 2 of the whole compact group of sequences. It's then standard that it's not Borel. Hence $L\cap P$ is not Borel and since $P$ is closed, it follows that $L$ is not Borel.
