Let $S$ be a measurable subset of $\Bbb{R^2}$. Assume for every $x\in S$ there exists a sequence of cubes $\{Q_k(x)\}$ centered at $x$ with side lengths tending to zero such that $$|S\cap Q_k(x) |\le\frac{1}{2}|Q_k(x)|$$ Show that $|S|=0$
I found that $\{Q_k(x)\}$ is a Vitali covering, so we have $|S\setminus\bigcup Q_j|=0$ and $\sum Q_j<(1+\epsilon)|S|$. Then $$|S|=|S\setminus \bigcup Q_j|+|\bigcup Q_j \cap S|=0+\sum |Q_j\cap S| \le \frac{1}{2}\sum |Q_j|$$
By $\sum Q_j<(1+\epsilon)|S|$ and let $\epsilon=\frac{1}{2}$, then $|S|<\frac{3}{4}|S|$, so contradiction. Thus $|S|=0$.
However, my teacher particularly pointed out that we can use the Lebesgue Differentiation theorem.
Could anyone kindly provide a hint on how to use the Lebesgue Diffrentiaion theorem here? Thanks!
Consider the function $\chi_S(x)$, the indicator function of $S$. By Lebesgue's differentiaton theorem,
$\lim_{|Q| \to 0} \frac{1}{|Q|} \int_Q \chi_S = \chi_S(x)$ for ae $x \in \mathbb{R}^2$.
Note that $\frac{1}{|Q|} \int_Q \chi_S = \frac{|Q \cap S|}{|Q|}$.
Suppose $|S| > 0$. Then we can find an $x \in S$ such that the conclusion of the Lebesgue differentiation theorem holds when applied to $\chi_S$ at $x$. That is,
$\lim_{|Q| \to 0} \frac{1}{|Q|} \int_Q \chi_s = 1$.
However, by assumption there's a sequence $Q_k$ of cubes whose sidelengths tend to $0$ with $|S \cap Q_k| \le \frac{1}{2} |Q_k|$.
From the above limit, choose $k$ sufficiently large so that $\frac{1}{|Q_k|} \int_{Q_k} \chi_S > \frac{1}{2}$.
Evaluating the integral as above and using the bound on $|S \cap Q_k|$ gives a contradiction.