Let $\mathbb{R}^d$ be a D816: Euclidean real Cartesian product.
Let $\mu^*$ be the D780: Lebesgue outer measure on $\mathbb{R}^d$.

Let $\mathbb{E} \subseteq \mathbb{R}^d$ and $x \in \mathbb{R}^d$. Let $\{ I_n \}_{n \in \mathbb{N}}$ be a countable covering of $E$ by open real $d$-intervals. Result R2974: Union of translated Euclidean real sets shows that then $\{ I_n + x \}_{n \in \mathbb{N}}$ is in turn a countable covering of $E + x$ by open real $d$-intervals. Results

(i)	R1052: Lebesgue outer measure coincides with the volume function on real n-intervals
(ii)	R1158: Translation invariance of Euclidean volume function
(iii)	R1049: Lebesgue outer measure is outer measure

now imply \begin{equation} \mu^*(E + x) \leq \mu^* \Big( \bigcup_{n \in \mathbb{N}} (I_n + x) \Big) \leq \sum_{n = 0}^{\infty} \mu^*(I_n + x) = \sum_{n = 0}^{\infty} \mathsf{Vol}(I_n + x) = \sum_{n = 0}^{\infty} \mathsf{Vol}(I_n) \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 + x) \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 set $E + x$ translated by $-x$ to obtain an inequality in the opposing direction \begin{equation} \mu^*(E) = \mu^*((E + x) - x) \leq \mu^*(E + x) \end{equation} An inequality in both directions was established, whence R1043: Equality from two inequalities for real numbers implies the equality $\mu^*(E + x) = \mu^*(E)$. This finishes the proof. $\square$