1.7

It suffices to show that $\forall \epsilon > 0, \exists \mathcal{O}$, open, such that $m^*(\mathcal{O}-\delta E) < \epsilon$. Before we move ahead, first define: $$\Delta = (\delta_1 \cdot \ldots \cdot \delta_d)$$ Recall that if $Q_k$ is a closed cube, $$Q_k = [a_1, b_1]\times[a_2,b_2]\times \ldots \times[a_d,b_d]$$ ...where $a_k < b_k \in \mathbb{R}$, and: $$m(Q_k) = \prod_{k = 1}^{d}(b_k - a_k)$$
$\dagger$ Observe that for any closed cube $Q$, $$\delta Q = [\delta_1 a_1, \delta_1 b_1]\times[\delta_2 a_2,\delta_2 b_2]\times \ldots \times[\delta_d a_d,\delta_d b_d]$$ $$\Rightarrow m(\delta Q) = \prod_{k = 1}^{d}\delta_k(b_k - a_k) = \Delta m(Q)$$
Next, observe that if $\mathcal{O}$ is open, $\delta \mathcal{O}$ is open. (This is somewhat trivial... just notice that $x \in \mathcal{O} \Rightarrow \exists r > 0$ s.t. $B_r{(x)} \subset \mathcal{O}$.

Now, just let $\tau = \min\left\lbrace \delta_1, \ldots, \delta_d \right\rbrace$, and observe that $x \in \mathcal{O} \Rightarrow \delta x \in \delta \mathcal{O}$ and $B_{\tau r}{(\delta x)} \subset \delta \mathcal{O}$.)

Next, since $E$ is measurable, $\forall \epsilon > 0, \exists \mathcal{O}$, open, such that $E \subset \mathcal{O}$, and $m^*(\mathcal{O}-E) < \frac{\epsilon}{\Delta}$. Certainly, $E \subset \mathcal{O} \Rightarrow \delta E \subset \delta \mathcal{O}.$

Lastly, notice: $$m^*(\delta \mathcal{O} - \delta E) = \inf \Big\lbrace \sum_{k=1}^{\infty} m(\delta Q_k) \hspace{0.25cm} \Big| \hspace{0.25cm} (\mathcal{O} - E) \subset \bigcup_{k=1}^{\infty} Q_k \Big\rbrace$$ ...which, by ($\dagger$): $$= \inf \Big\lbrace \Delta \sum_{k=1}^{\infty} m(Q_k) \hspace{0.25cm} \Big|\hspace{0.25cm} (\mathcal{O} - E) \subset \bigcup_{k=1}^{\infty} Q_k \Big\rbrace = \ldots$$
$$\ldots = \Delta m^*(\mathcal{O}-E) < \Delta \frac{\epsilon}{\Delta} = \epsilon$$ ...as desired.