ThmDex – An index of mathematical definitions, results, and conjectures.
Reflection invariance of Lebesgue outer measure
Formulation 0
Let $\mathbb{R}^d$ be a D816: Euclidean real Cartesian product.
Let $\mu^*$ be the D780: Lebesgue outer measure on $\mathbb{R}^d$.
Then \begin{equation} \forall \, E \subseteq \mathbb{R}^d : \mu^*(-E) = \mu^*(E) \end{equation}
Proofs
Proof 0
Let $\mathbb{R}^d$ be a D816: Euclidean real Cartesian product.
Let $\mu^*$ be the D780: Lebesgue outer measure on $\mathbb{R}^d$.
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
(i) R1052: Lebesgue outer measure coincides with the volume function on real n-intervals
(ii) R1160: Reflection invariance of Euclidean volume function
(iii) R1049: Lebesgue outer measure is outer measure

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. $\square$