Exercise 1.6.25 (Steinhaus theorem II)

Question

Let $E \subset {\bf R}^d$ be a measurable set of positive measure, and let $\varepsilon > 0$.
\begin{enumerate}[label=(\roman*)]
\item Using \href{./exercise_1-6-15}{Exercise 1.6.15} and \href{./exercise_1-6-24}{Exercise 1.6.24}, show that there exists a cube $Q \subset {\bf R}^d$ of positive sidelength such that $m(E \cap Q) > (1-\varepsilon) m(Q)$.
\item Give an alternate proof of the above claim that avoids the Lebesgue differentiation theorem.
\item Use the above result to give an alternate proof of the Steinhaus theorem (\href{./exercise_1-6-8}{Exercise 1.6.8}).
\end{enumerate}

Elu Thingol · Accepted Answer

\textbf{(i)} \begin{proof} By \href{./exercise_1-6-15}{Exercise 1.6.15}, for every point $x \in E$, we have $$\lim_{r \to 0} \dfrac{1}{m\left(Q_r\right)} \int_{x+Q_r} 1_E = 1_E(x)$$ whenever $Q$ is a measurable subset of $B(0,r)$ with the property that $m(Q_r) \geq cm(B(0,r))$ for some $c>0$ independent of $r$. By \href{./exercise_1-6-24}{Exercise 1.6.24}, for almost every $x \in E$, the right-hand side term is one, i.e., $$\lim_{r \to 0} \dfrac{1}{m\left(Q_r\right)} \int_{x+Q_r} 1_E = 1.$$ Since $E$ is assumed to have a positive measure, we can safely fix a $x \in E$ satisfying this property. By the definition of limit, we can find a $r>0$ such that eventually $$1 - \dfrac{1}{m\left(Q_r\right)} \int_{x+Q_r} 1_E < \epsilon$$ holds for the $\epsilon > 0$ of our choice. In other words, $$1 - \dfrac{m(E \cap Q_r')}{m\left(Q_r'\right)} < \epsilon.$$ where $Q_r'$ is simply the translate $x + Q_r$. Rearranging the terms, we obtain $$m(Q_r')(1-\epsilon) < m(E \cap Q_r').$$ In particular, this is true of a cube $Q_r := \left[-\dfrac{r}{\sqrt{d}},\dfrac{r}{\sqrt{d}}\right]^d$, which (1) is inscribed inside the ball $B(0,r)$ and (2) exceeds the volume of a ball when multiplied by some constant independent of $r$. \begin{figure} \includegraphics[width=0.3\textwidth]{tao_measure_exercise_1-6-25_circle-square.png} \includegraphics[width=0.3\textwidth]{tao_measure_exercise_1-6-25_sphere-cube.png} \caption{A maximal square inscribed into a circle and a maximal cube inscribed into a sphere. If the main diagonal of the cube is to be equivalent to the ball diameter of $2r$, then the sides must be of length $\sqrt{2}r$ or $2/\sqrt{3}r$ respectively.} \end{figure} It is easy to verify both properties. We have $$\sup_{x \in Q_r} \|x\| = \sup_{x \in Q_r} \sqrt{\sum_{i=1}^{d}x_i^2} = \sqrt{\sum_{i=1}^{d} \left(\frac{r}{\sqrt{d}}\right)^2} = \sqrt{r^2} = r.$$ Thus, $Q_r \subseteq B(0,r)$. Furthermore, recall from \href{./exercise_1-1-10}{Exercise 1.1.10} that the volume of $B(0,r)$ can be expressed as $c_d \cdot r^d$ where $c_d$ is some constant independent of $r$. Since the volume of $Q_r$ is $(2r/\sqrt{d})^d$, we must require that for all $r>0$: $$c \left(\dfrac{2r}{\sqrt{d}}\right)^d \geq c_d r^d \iff c \geq \dfrac{c_d \sqrt{d}^d}{2^d}.$$ Fixing some $c$ satisfying the above inequality will yield the desired property $m(Q_r) \geq cm(B(0,r))$. Setting $Q'_r := x + Q_r$ finishes the proof. \end{proof} ____________________ \newline \textbf{(ii)} \begin{proof} %We first start with a simpler case when $E$ is bounded. By \href{./exercise_1-2-16}{Exercise 1.2.16 (Criteria for finite measure)}, we can find an integer $n$ and a finite union $F_n = \bigcup Q_i$ of closed dyadic cubes of sidelength $2^{-n}$ such that $m(E \triangle F_n) \leq \epsilon$. Pick an arbitrary dyadic cube $Q_n$ out of the dyadic cubes that constitute for the union $F_n$. From elementary set-theoretic identities we deduce %\begin{align*} %m\left(Q_n\right) - m\left(E \cap Q_n\right) %&= m\left(Q_n \cap E\right) + m\left(Q_n \setminus E\right) - m\left(E \cap Q_n\right)\ %&= m\left(Q_n \setminus E\right) \leq m\left(F_n \setminus E\right) \leq \epsilon %\end{align*}. %Thus, %\begin{align*} %1 - \dfrac{m\left(E \cap Q_n\right)}{m(Q_n)} %&\leq 1 - \dfrac{m\left(E \cap Q_n\right)}{m\left(E \cap Q_n\right)+\epsilon}\[5pt] %&= \dfrac{\epsilon}{m\left(E \cap Q_n\right)+\epsilon}\[5pt] %&\leq \dfrac{\epsilon}{m\left(E\right)+\epsilon} %\end{align*} %Multiplying both sides with $m(Q_n)$ we obtain %$$m\left(E \cap Q_n\right) > (1-\epsilon) m(Q).$$ %Should $E$ be unbounded, apply the above to $E \cap B(0,r)$ for $r>0$ large enough. \end{proof} ____________________\newline \textbf{(iii)} \begin{proof} \end{proof}

Exercise 1.6.25 (Steinhaus theorem II)

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Exercise 1.6.25 (Steinhaus theorem II)

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