1.27

Since $E_1$ and $E_2$ are compact sets in $\mathbb{R}^d$, certainly $E_2 \backslash E_1$ is a bounded and measurable. For any $t > 0$, The sets $(E_2 \backslash E_1) \cap \overline{B_{t}(0)}$ are also bounded and measurable. For simplicity's sake, from here on out, let's say:

$$S_t = (E_2 \backslash E_1) \cap \overline{B_{t}(0)}$$ Define $f(t) = m(S_t)$. If I can prove $f$ is a continuous function, we're done --(by the intermediate value theorem).

Side note: Notice the function $g(t) = \alpha t^d$ is continuous, but not uniformly continuous. ($\dagger$)

Let $0 \leq \tau < t$. Notice $|f(t) - f(\tau)| = |m(S_t) - m(S_\tau)|$.

Certainly, $S_\tau \subset S_t$, thus:.

$$|m(S_t) - m(S_\tau)| = m(S_t) - m(S_\tau) = m(S_t \backslash S_\tau)$$ Now, notice that $$(S_t \backslash S_\tau) \subset \overline{B_t(0)} \backslash \overline{B_\tau(0)}$$ so: $$m(S_t \backslash S_\tau) \leq m(\overline{B_t(0)} \backslash \overline{B_\tau(0)}) \leq \alpha(d)\big(t^d - \tau^d)$$ where $\alpha(d) = \frac{\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2} + 1)}$, the volume of the $d$-dimensional unit ball. (Recall we proved the relation between $B_1$ and $B_t$ in problem 6 of this chapter.) Thus, by $\dagger$, $$|t - \tau| < \delta \Rightarrow \alpha(d)\big(t^d - \tau^d) \leq \epsilon(t)$$ Therefore, $f$ is a continuous function, and by the IVT, given any $c$ with $a < c < b$, we can find a $t^* > 0$ such that $m(E) = m\big(E_1 \cup S_{t^*}\big) = c$, where $E_1 \cup S_{t^*}$ is compact (by construction).