Let $E \subseteq \mathbb{R}^d$. Let $\{ I_n \}_{n \in \mathbb{N}}$ be a countable covering of $E$ by open real $d$-intervals. Result
R2985: Set covering of Euclidean real reflected set says that then $\{ - I_n \}_{n \in \mathbb{N}}$ is in turn a countable covering of the reflected set $- E$ by open real $d$-intervals. Results
now imply
\begin{equation}
\begin{split}
\mu^*(- E) \leq \mu \Big( \bigcup_{n \in \mathbb{N}} - I_n \Big) \leq \sum_{n \in \mathbb{N}} \mu^*(- I_n) = \sum_{n \in \mathbb{N}} \mathsf{Vol}(- I_n) = \sum_{n \in \mathbb{N}} \mathsf{Vol}(I_n)
\end{split}
\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^*(- E) \leq \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) = \mu^*(E)
\end{equation}
We may now proceed to apply this inequality in turn to the reflected set $-E$ itself to deduce
\begin{equation}
\mu^*(E) = \mu^*(- (-E)) \leq \mu^*(- E)
\end{equation}
An inequality in both directions was established, whence
R1043: Equality from two inequalities for real numbers implies the equality $\mu^*(- E) = \mu^*(E)$. This finishes the proof. $\square$