From "An Introduction to Lebesgue Integration and Fourier Series" by Howard J. Wilcox and David L. Myers:
$E$ is the unit interval $[0,1]$.
7.4 Definition: The outer measure $m^{*}(A)$ of any set $A \subset E$ is defined to be the real number $\text{glb} \{m^{*}(G) \mid \text{$A \subset G$ and $G$ open in $E$}\}$.
7.6 Lemma: Let $A \subset E$. Then for any $x$, $m^{*}(A) = m^{*}(x + A)$, 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^{*}(x + A) &= \text{glb} \{m^{*}(G) \mid \text{$x + A \subset G$ and $G$ open in $E$}\} \\ &= \text{glb} \{m^{*}(-x + G) \mid \text{$A \subset -x + G$ and $-x + G$ open in $E$}\} \\ &\geq m^{*}(A). \end{align*}
The proof that $m^{*}(A) \geq m^{*}(x + A)$ is similar.
I am uncertain as to the reason for the approach to the last step. It seems that $$ \{m^{*}(-x + G) \mid \text{$A \subset -x + G$ and $-x + G$ open in $E$}\} = \{m^{*}(G) \mid \text{$A \subset G$ and $G$ open in $E$}\} $$
and therefore \begin{align*} \text{glb} \{m^{*}(-x + G) \mid \text{$A \subset -x + G$ and $-x + G$ open in $E$}\} &= \text{glb} \{m^{*}(G) \mid \text{$A \subset G$ and $G$ open in $E$}\} \\ &= m^{*}(A). \end{align*}
Is this correct?