Stein 1.7

$\textbf{Exercise 1.7}$ Let $\delta=(\delta_1\delta_2 \cdots \delta_d)$ be a $d-$tuple of positive real numbers. Let $E \in{\mathcal{M}}$ and define $$\delta E = \{(\delta_1x_1,...,\delta_dx_d) : (x_1,...,x_d) \in{E} \}$$ So, $\delta E$ is a dilation of the set $E$ by an amount of $\delta_j$ along each $j$-axis of $\mathbb{R}^d$. (This notation suggests that we are using the standard basis for $\mathbb{R}^d$).
Show that $\delta E \in{\mathcal{M}}$ and that $m(\delta E) = \delta_1 \cdots \delta_d \cdot m(E)$

$\textit{Proof}: $
$E \in{\mathcal{M}}$ means that $\forall \epsilon >0$, $\exists \mathcal{O} \supseteq E$, $\mathcal{O}$ open where $m_*(\mathcal{O}-E) \leq \epsilon$. By definition, $m_*(\mathcal{O}-E)= \inf \sum_{j=1}^\infty |Q_j|$ where the infimum is taken over all countable coverings of closed cubes, $\mathcal{O}- E \subseteq \bigcup_{j=1}^\infty Q_j$. Since the infimum of a sequence of real numbers is, by definition, an accumulation point, this means that there exists a sufficiently close covering of $\mathcal{O}-E$ s.t. $\sum_{j=1}^\infty |Q_j| \leq \epsilon $. Since $E$ has a covering by cubes, then $\delta E$ has a covering by rectangles, which can be approximated by cubes. So, with $$\mathcal{O}-E \subseteq \bigcup_{j=1}^\infty Q_j \quad \text{then} \quad \delta(\mathcal{O}-E)=\delta \mathcal{O}-\delta E \subseteq \bigcup_{j=1}^\infty \delta Q_j$$ Using the monotinicity and sub-additivity of $m_*$ $$m_*(\delta \mathcal{O}-\delta E) \leq m_*\left( \bigcup_{j=1}^\infty \delta Q_j \right) \leq \sum_{j=1}^\infty m_*(\delta Q_j) = \sum_{j=1}^\infty |\delta Q_j|$$ Since the side lengths of cubes align with the respective coordinate axes in $\mathbb{R}^d$, the $\delta$-dilation of a cube is known: $|\delta Q_j|=|\delta_1 \cdots \delta_d| \cdot |Q_j|$. So, $$\cdots = \sum_{j=1}^\infty |\delta_1 \cdots \delta_d| \cdot | Q_j| = |\delta_1 \cdots \delta_d| \sum_{j=1}^\infty |Q_j| \leq \delta_1 \cdots \delta_d \cdot \epsilon $$ Thus, $\delta E \in{\mathcal{M}}$. Also, $$m(\delta E) = m_*(\delta E) = \inf \sum_{j=1}^\infty |\delta Q_j| = \delta_1 \cdots \delta_d \cdot \inf \sum_{j=1}^\infty |Q_j|=\delta_1 \cdots \delta_d \cdot m_*(E) = \delta_1 \cdots \delta_d \cdot m(E)$$