Suppose for every $x \in [0, 1]$, we have a subset $S_x$ of the natural numbers with asymptotic density $1$ such that if $n \in S_x$, there is an open neighbourhood $U$ of $x$ (depending on $x$ and $n$) such that $n \in S_y$ for all $y \in U$.

**Question:**

For any $\varepsilon > 0$ can I find a subset $M$ of $[0, 1]$ with $\mu(M) > 1 - \varepsilon$, and a subset $K \subset \mathbb N$ of positive upper density satisfying the following condition?

- For all $x \in M$, there exists some $N > 0$ such that $K \cap [N, \infty)$ is a subset of $S_x \cap [N, \infty)$.

*A word on motivation:* This question arose in trying to prove this
result on sequences of “almost equicontinuous” functions. The result as stated turns out to be false as shown by Mateusz Kwaśnicki in the comments and Iosif Pinelis in his answer, but I believe if this lemma is true, I can get a subsequence that converges in measure to *some* function $f$, not necessarily continuous.