1) Problem: Prove that for any non-empty set $S \subseteq \mathbb{R}$ and $a > 0$, the Lebesgue outer measure satisfies $$m^*(aS) = a m^*(S).$$ 2) Formula and definitions: - The Lebesgue outer measure $m^*(S)$ is defined as the infimum of the sums of lengths of open intervals covering $S$. - For $a > 0$, the set $aS = \{t \in \mathbb{R} : t = a x \text{ for some } x \in S\}$. 3) Proof steps: Step 1: Let $\{I_n\}$ be a countable collection of open intervals covering $S$, so $S \subseteq \bigcup_n I_n$ and $$m^*(S) \leq \sum_n |I_n|,$$ where $|I_n|$ is the length of interval $I_n$. Step 2: Consider the intervals $a I_n = \{a x : x \in I_n\}$. Since $a > 0$, the length of $a I_n$ is $$|a I_n| = a |I_n|.$$ Also, $a S \subseteq \bigcup_n a I_n$. Step 3: By definition of outer measure, $$m^*(a S) \leq \sum_n |a I_n| = \sum_n a |I_n| = a \sum_n |I_n|.$$ Step 4: Taking the infimum over all such covers $\{I_n\}$ of $S$, we get $$m^*(a S) \leq a m^*(S).$$ Step 5: To prove the reverse inequality, apply the same argument to $S = \frac{1}{a} (a S)$ with $\frac{1}{a} > 0$: $$m^*(S) = m^*\left(\frac{1}{a} (a S)\right) \leq \frac{1}{a} m^*(a S) \implies a m^*(S) \leq m^*(a S).$$ Step 6: Combining both inequalities, $$m^*(a S) = a m^*(S).$$ Final answer: $$\boxed{m^*(a S) = a m^*(S)}.$$