# Outer measure of translate of set

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?