Ex3-18

Question. Prove that the outer measure is translation invariant; that is, if E_{\mathbf{h}} = \{\mathbf{x} + \mathbf{h} \in \R^n \mid \mathbf{x} \in E\} is the translate of E by \mathbf{h} \in \R^n, show that |E_\mathbf{h}|_e = |E|_e. If E is measurbale, show that E_{\mathbf{h}} is also measurable. 
Answer. Given a countable cover S = \{I_i\}_{i \in \Z^+} of E by closed intervals, observe that S_{\mathbf{h}} = \left\{(I_i)_\mathbf{h}\right\}_{i \in \Z^+} is a countable cover of E_\mathbf{h} by closed intervals, where (I_i)_{\mathbf{h}} is I_i translated by \mathbf{h} for each i \in \Z^+. Clearly \sigma(S) = \sigma(S_\mathbf{h}) since v(I_i) = v((I_i)_\mathbf{h}). Thus, 
\begin{align}
\mathcal{S} = \{\sigma(S) \mid S \ & \text{is a countable cover of } E \text{ by closed intervals}\} \\
&\sube \{\sigma(S_\mathbf{h}) \mid S_{\mathbf{h}} \text{ is a countable cover of } E_\mathbf{h} \text{ by closed intervals}\} = \mathcal{S}_\mathbf{h},
\end{align} 
so
\inf(\mathcal{S}) \ge \inf(\mathcal{S}_\mathbf{h}).
Repeating the argument for E_\mathbf{h} and (E_\mathbf{h})_{-\mathbf{h}} = E gives us the opposite inequality, therefore,
|E|_e = \inf(\mathcal{S}) = \inf(S_\mathbf{h}) = |E_\mathbf{h}|_e,
which proves the first part.
 
Now, suppose that E is measurable. Then for every \varepsilon > 0, there exists an open set G with E \sube G such that |G \setminus E|_e < \varepsilon. Then G_\mathbf{h} is an open set such that 
|G_\mathbf{h} \setminus E_\mathbf{h}| _ = |(G \setminus E)_\mathbf{h}|_e = |G \setminus E| < \varepsilon. 
This proves that E_\mathbf{h} is also measurable.