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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?

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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.

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  • $\begingroup$ Thanks for your answer. One more question: may you explain why $\Phi(L\cap P)$ is the set of sequence in $\{0,1\}^{\mathbf{N}}$ which are $\omega$-convergent to $0$? $\endgroup$
    – Paolo Leonetti
    Commented Jan 15, 2019 at 9:57
  • $\begingroup$ Because for $a\in P$, $\frac{1}{u_n}\sum_{i=1}^{u_n}a_i$ is equal to $a_{u_n}+o(1)$. $\endgroup$
    – YCor
    Commented Jan 16, 2019 at 1:02
  • $\begingroup$ Yes, this is clear. However, assume that $v(\mathscr{F})\text{-}\lim a_{u_n}=0$, i.e., $\{n \in \mathbf{N}: a_{u_n}=0\}=v(F)$, for some $F \in \mathscr{F}$ ($\star$). Then it is claimed that $\mathscr{F}\text{-}\lim \sum_{i\le n}\frac{a_i}{n}=0$, that is, $$\left\{n\in \mathbf{N}: \sum_{i\le n}\frac{a_i}{n} \le \varepsilon \right\} \in \mathscr{F}$$ for all $\varepsilon>0$. How does it follow from ($\star$)? $\endgroup$
    – Paolo Leonetti
    Commented Jan 16, 2019 at 13:28
  • $\begingroup$ @PaoloLeonetti In my previous comment I have two sequences differing by $o(1)$. So if one $\mathscr{F}$-converges to 0, so does the other. $\endgroup$
    – YCor
    Commented Jan 16, 2019 at 13:46
  • $\begingroup$ Sorry for the delay. You have $$ \omega\text{-}\lim a_{u_n}=0\,\,\,\,\text{ if and only if }\,\,\,\,\omega\text{-}\lim \frac{1}{u_n}\sum_{i\le u_n}a_i=0, $$ this is clear. Here $\omega$ is the image of $\mathscr{F}$ through $v$, which should be defined as $\{S\subseteq \mathbf{N}: v^{-1}(S) \in \mathscr{F}\}$. However, I still don't get why $$ \Phi(L\cap P)=\{a \in \{0,1\}^{\mathbf{N}}: \omega\text{-}\lim a_n=0\}. $$ Am I missing something easy? $\endgroup$
    – Paolo Leonetti
    Commented Jul 23, 2019 at 18:19

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