a sequence defined by a measurable set, with Lebesgue measure 1/2

Question

Let $A$ be a subset of $[0,1]$ with Lebesgue measure $\mu(A) = 1/2$.
Let $I_{i,n} ~ = ~ [(i-1)/n,i/n]$.

I am interested in the sequence: $$S_n(A) = n\sum_{i=1}^n \mu(A \cap I_{i,n} )^2 $$

I can show that $S_n(A)$ is between $1/4$ and $1/2$.

My question is: Does $S_n(A)$ converge to $1/2$ ?

I can show it is true when $A$ is a finite union of intervals. For large $n$, all except a finite number of the $I_{i,n}$ are inside or outside $A$. For general $A$, the Lebesgue density theorem seems relevant.

If the answer is no, then what is the smallest possible value of $\liminf S_n(A)$, as $A$ ranges over all measurable sets with $\mu(A) = 1/2$ ?

Answer 1

The answer to the question is yes.

Rewrite the sequence as: $$S_n(A) = \sum_{i=1}^n n^{-1} ( n \mu(A \cap I_{i,n} ) )^2 $$ For each $n$ define a step function by: $$f_n(x) = n \mu( A \cap I_{i,n} ) ~~ {\rm for} ~~ x \in I_{i,n}$$ and note that $$\int_0^1 f_n(x)^2 \,dx = S_n(A) $$ Fix an $x \in [0,1]$ and set $E_n$ to the interval $I_{i,n}$ that contains $x$. Then $E_n$ "shrinks nicely" to $x$ (as defined by Rudin). By the Lebesgue differentiation theorem, $f_n$ converges a.e. to ${\bf1}_A$ $$f_n \rightarrow {\bf1}_A ~~~ \rm a.e.$$ where ${\bf1}_A$ is the indicator function of $A$. Because squaring is continuous $$f_n^2 \rightarrow {\bf1}^2_A = {\bf1}_A ~~~ \rm a.e.$$

Obviously $|f_n^2| \le {\bf1}$, so by the Lebesgue dominated convergence theorem $$\int_0^1 f_n^2 \,dx \rightarrow \int_0^1 {\bf1}_A \,dx ~~=~~ \mu(A) = 1/2$$

As a followup, does anyone know of a set $A$ where the difference between these integrals goes to $0$ slower than $O(1/n)$ ? For example, $O(1/\sqrt n )$ ? Terry Tao has written that the convergence can be arbitrarily slow, but I am not able to make an example.