Existence of sequence of Borel subsets of the unit interval

Question

Consider the probability space $([0,1], \mathscr{B}([0,1]), \lambda)$, where $\lambda$ is the Lebesgue measure.

Question. Does there exist a sequence $(S_n)$ of Borel subsets of $[0,1]$ with the property that $\lim_n \lambda(S_n)=0$ and $$ \limsup_{N\to \infty}\frac{| \{n \in \{1,\ldots,N\}: x \in S_n \}|}{N}>0 $$ for $\lambda$-almost all $x \in [0,1]$?

Replacing the centered property with the weaker condition "$\exists^\infty n, x \in S_n$", the answer is affirmative by setting, e.g., $(S_n)=([0,1], [0,1/2], [1/2,1], [0,1/3], [1/3, 2/3],...)$.

Answer 1

We can use the same sets as in the weaker condition you cited as long as we repeat each set in the sequence enough times. Suppose you want to have the limsup at least $\frac{3}{4}$ for all $x$ (any desired limsup in $(0, 1)$ will work).

Let $S_1 = [0, 1]$. With $N = 1$, $$ \frac{|\{n \in \{1, \dots, N\} : x \in S_n\}|}{N} = 1 \quad\text{for all }x \in [0, 1]. $$ So far, so good. We can also keep $S_2 = [0, \frac{1}{2}]$. Now, for $N = 2$, $$ \frac{|\{n \in \{1, \dots, N\} : x \in S_n\}|}{N} = 1 \quad\text{for all }x \in [0, \frac{1}{2}]. $$ If $S_3 = [\frac{1}{2}, 1]$, then for $N = 3$, we only have $$ \frac{|\{n \in \{1, \dots, N\} : x \in S_n\}|}{N} = \frac{2}{3} \quad\text{for all }x \in [\frac{1}{2}, 1]. $$ That is unacceptable. To keep the limsup high for these values of $x$, we'll set $S_4 = S_3$. Now for $N = 4$, we have $$ \frac{|\{n \in \{1, \dots, N\} : x \in S_n\}|}{N} = \frac{3}{4} \quad\text{for all }x \in [\frac{1}{2}, 1]. $$ While we've been giving higher values of $x$ more sets, the lower values have been losing out. Thus, when we set $S_5 = [0, \frac{1}{3}]$, for $N = 5$, we have $$ \frac{|\{n \in \{1, \dots, N\} : x \in S_n\}|}{N} = \frac{3}{5} \quad\text{for all } x \in [0, \frac{1}{3}] $$ This is not quite high enough, so we set $S_8 = S_7 = S_6 = S_5$ to ensure that limsup stays over $\frac{3}{4}$ for these values of $x$.

We can continue in this way, repeating sets in the original sequence enough times (the number of repetitions probably grows exponentially?) that the limsup will be at least $\frac{3}{4}$ for every $x \in [0, 1]$. And the measure of the sets goes to zero, very slowly.