Show that the Lebesgue Outer Measure is Translation Invariant I am currently proving the following theorem in my measure theory course:

Let $A \in \mathcal P(\mathbb{R})$ and $t \in \mathbb{R}$. Then $m^*(A+t)=m^*(A)$, where $m^*$ is the Lebesgue outer measure.

Here is the first half of my proof and also where I am unsure of a few details.
Define a sequence of open intervals $(I_n)_{n=1}^\infty$ that covers $A$. Therefore, $(I_n+t)_{n=1}^\infty$ is a sequence of open intervals that covers $A+t$. So we can see that $$m^*(A+t) \leq \sum_{n=1}^{\infty}l(I_{n}+t)=\sum_{n=1}^{\infty}l(I_{n})$$ Where $l(I_n)$ is the length of the intervals. So for every sequence of open intervals $(I_n)_{n=1}^\infty$ that cover $A$ we have $m^*(A+t) \leq \sum_{n=1}^{\infty}l(I_{n})$, therefore $m^*(A+t) \leq m^*(A)$.

I believe everything is correct but my reasoning for why we can conclude that  $m^*(A+t) \leq m^*(A)$. Is this due to the monotonicity and sub-additivity of $m^*$? More precisely, is it because $$m^*(A+t) \leq m^*(A) \leq m^*(\bigcup_{n \geq1}I_{n})\leq \sum_{n=1}^{\infty}l(I_{n})$$ since $A \subseteq \bigcup_{n \geq1}I_{n}$ and we concluded that $\sum_{n=1}^{\infty}l(I_{n}+t)=\sum_{n=1}^{\infty}l(I_{n})$?
Any feedback is welcome.
 A: You wrote

So for every sequence of open intervals $(I_n)_{n=1}^\infty$ that cover $A$ we have $m^*(A+t) \leq \sum_{n=1}^{\infty}l(I_{n})$, therefore $m^*(A+t) \leq m^*(A)$.

This follows immediately by definition of the outer measure as an infimum of a set of real numbers. Recall that if $E\subset \Bbb{R}$ is non-empty, then $\inf E$ is by definition an element of $[-\infty,\infty)$ such that two things are satisfied:

*

*$\inf E$ is a lower bound for the set $E$: for every $x\in E$, we have $\inf E \leq x$.

*For any lower bound $\beta$ of $E$, we have $\beta \leq \inf E$.

In your case, we're fixing a set $A\subset\Bbb{R}$, and letting $E$ be the set of all sums $\sum_{n\geq 1} l(I_n)$, where $\{I_n\}_{n=1}^{\infty}$ is a countable cover for $A$ consisting entirely of open intervals. What your proof shows is that $m^*(A+t)$ is a lower bound for $E$, hence by definition of infimum, $m^*(A+t)\leq \inf E =: m^*(A)$.
A: How do you deine $m^*$?
If your definition is:
$$m^*(A)=\inf\{\sum_n l(J_n):\ A\subset\bigcup_n J_n,\, J_n\text{ are open intervals}\}$$
then you have proven that given ANY open cover (by intervals) of $A$,
$m*(A+t)\le \sum_n l(I_n)$.
You only have to take $\inf$ over all possible covers to get $m^*(A+t)\le m^*(A)$.

The reverse inequality is a consequence of the previous inequality:
$m^*(A)=m^*((A+t)-t)\le m ^*(A+t)$
