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From "An Introduction to Lebesgue Integration and Fourier Series" by Howard J. Wilcox and David L. Myers:

For the remainder of this chapter, we will denote the unit interval $[0,1]$ by $E$.

7.2 Theorem: Every non-empty open set $G \subset \mathbb{R}$ can be expressed uniquely as a finite or countably infinite union of pairwise disjoint open intervals.

7.3 Definition: The outer measure $m^{\star}(G)$ of an open set $G \subset E$ is defined as the real number $\sum_{i} (b_{i} - a_{i})$, where $G = \bigcup_{i} (a_{i}, b_{i})$ as in Theorem 7.2.

7.4 Definition: The outer measure $m^{\star}(A)$ of any set $A \subset E$ is defined to be the real number $\text{glb } \{ m^{\star}(G) \mid A \subset G \text{ and } G \text{ open in } E \}$.

7.6 Lemma: Let $A \subset E$. Then for any $x$, $m^{\star}(A) = m^{\star}(x + A)$, where $x + A = \{ x + a \mid a \in A \}$ is called the translate of the set $A$ by $x$. (Since we are restricted to subsets of $E$, we may need to translate modulo 1; for example, $[1/2, 1] + 1/4 = [3/4, 1] \cup [0, 1/4]$.)

Proof: If $A$ is an interval or a countable union of pairwise disjoint open intervals, the lemma is clearly true. Thus the lemma holds if $A$ is any open set in $E$. For arbitrary $A \subset E$,

\begin{align} m^{\star}(x + A) &= \text{glb } \{ m^{\star}(G) \mid x + A \subset G \text{ and } G \text{ open in } E \} \\ &= \text{glb } \{ m^{\star}(-x + G) \mid A \subset -x + G \text{ and } -x + G \text { open in } E \} \\ &\geq m^{\star}(A). \end{align}

The proof that $m^{\star}(A) \geq m^{\star}(x + A)$ is similar.

How can I show that \begin{align} & \text{glb } \{ m^{\star}(G) \mid x + A \subset G \text{ and } G \text{ open in } E \} \\ & = \text{glb } \{ m^{\star}(-x + G) \mid A \subset -x + G \text{ and } -x + G \text { open in } E \}? \\ \end{align}

Why not let $H = -x + G$ to obtain $\text{glb } \{ m^{\star}(H) \mid A \subset H \text{ and } H \text { open in } E \} = m^{\star}(A)$ instead of $\text{glb } \{ m^{\star}(-x + G) \mid A \subset -x + G \text{ and } -x + G \text { open in } E \} \geq m^{\star}(A)$?

Why does $\text{glb } \{ m^{\star}(-x + G) \mid A \subset -x + G \text{ and } -x + G \text { open in } E \} \geq m^{\star}(A)$ hold?

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