Let $E \subseteq \mathbb{R}^d$. Let $\{ I_n \}_{n \in \mathbb{N}}$ be a countable covering for $E$ of open real $d$-intervals. Result
R2987: Set covering of Euclidean real scaled set shows than then the collection of scaled intervals $\{ \lambda I_n \}_{n \in \mathbb{N}}$ is in turn a countable covering for $\lambda E$ of open real $d$-intervals. Results
now imply
\begin{equation}
\mu^*(\lambda E) \leq \mu^* \Big( \lambda \bigcup_{n \in \mathbb{N}} I_n \Big) \leq \sum_{n \in \mathbb{N}} \mu^*(\lambda I_n) = \sum_{n \in \mathbb{N}} \mathsf{Vol}(\lambda I_n) = |\lambda|^d \sum_{n \in \mathbb{N}} \mathsf{Vol}(I_n)
\end{equation}
Since $\{ I_n \}_{n \in \mathbb{N}}$ was an arbitrary covering of $E$, we may apply
R1109: Antitonicity of infimum to extend the above inequality to the infimum element on the right hand side over all such coverings of $E$ to obtain
\begin{equation}
\mu^*(\lambda E) \leq |\lambda|^d \inf_{\substack{J_0, J_1, J_2, \dots \in \mathsf{POI}(\mathbb{R}^d) : \\ E \subseteq \bigcup_{n \in \mathbb{N}} J_n}} \sum_{n \in \mathbb{N}} \mathsf{Vol}(J_n) = |\lambda|^d \mu^*(E)
\end{equation}
We may now proceed to apply this inequality in turn to the scaled set $\frac{1}{\lambda} E$ to obtain an inequality in the opposing direction
\begin{equation}
\mu^*(E) = \mu^* \Big( \lambda \Big( \frac{1}{\lambda} E \Big) \Big) = \mu^* \Big( \frac{1}{\lambda} (\lambda E) \Big) \leq \frac{1}{|\lambda|^d} \mu^*(\lambda E)
\end{equation}
Multiplying both sides by the nonzero quantity $|\lambda|^d$ yields now $|\lambda|^d \mu^*(E) \leq \mu^*(\lambda E)$. An inequality in both directions was established, whence
R1043: Equality from two inequalities for real numbers implies the equality $\mu^*(\lambda E) = |\lambda|^d \mu^*(E)$. This finishes the proof. $\square$
