Prove $m^*(A) = m^*(A + t)$ Define $m^*(A) = \inf Z_A$ as the outer measure of $A \subseteq \mathbb{R}$ where
$$Z_A = \left\{\sum_{n=1}^{\infty}|I_n| : I_n  \text{   are intervals}, A \subseteq \bigcup_{n=1}^{\infty}I_n\right\}  $$

We want to show $m^*(A) = m^*(A + t)$. It will suffice to show $Z_A = Z_{A + t}$. So, pick an element $z \in Z_A$. Therefore, we have a covering $\{I_n\}$ of $A$ such that $z = \sum |I_n |$. Let $I_n = [a_n, b_n]$. then $|I_n| = b_n - a_n = (b_n + t) - (a_n + t) = |I_n + t|$, and also since $\{I_n + t\}$ cover $A +t$, then $z \in Z_{A + t}$.
Now, pick $z \in Z_{A + t}$. So, we have a covering $\{J_n\}$ of $A + t$ such that $z = \sum |I_n + t |$. But, notice $|I_n + t| = |I_n|$ by the above computation. Also, we have that $\{J_n - t \}$ cover $A$. Therefore, $z \in Z_A$

Is this proof correct? is this rigorous enough to conclude that since $(I_n)$ cover $A$, then $(I_n + t)$ must cover $A + t$ and reverse? Thanks for the feedback.

You may want to explain why it suffices to show $Z_A = Z_{A+t}$.
The argument for $Z_A \subseteq Z_{A+t}$ looks good (you may want to use the subset notation to make what you're doing clearer, but I don't think it's necessary).
The argument for $Z_{A+t} \subseteq Z_A$ uses both $J_n$ and $I_n$; I think you should just be referring to $J_n$, or if you prefer, explicitly stating how the collections $\{J_n\}$ and $\{I_n\}$ are related.
After a little bit of tidying up, it looks like it will be a good proof.

Note, there is an easier way to show $Z_{A + t} \subseteq Z_A$. You've shown $Z_A \subseteq Z_{A + t}$ for any real number $t$. Replace $t$ by $s$ so that $Z_A \subseteq Z_{A+s}$. Now replacing $A$ by $A + t$ we have $Z_{A+t} \subseteq Z_{A+t+s}$. Choosing $s = -t$ we have the desired result.
