the Lebesgue outer measure of an n-interval equals its geometric measure I'm trying to prove that $\lambda^*(I)=vol(I)$ for an arbitrary $n$-interval $I\subset \mathbb{R}^n$, where $\lambda^*(I)$ is the Lebesgue outer measure on $\mathbb{R}^n$ and $vol(I)$ is the geometric measure of $I$. The proof is quite clear in the case that $I$ is not bounded so i will assume that $I$ is bounded. Here's my attempt:
We will first show that $\lambda^*(I)\ge vol(I)$. Choose a collection $\{I_i^*\}_{i\in\mathbb{N}_1}$ of intervals such that
$$ I\subset \bigcup_i I_i^*.$$
Now by the monotonicity of the geometric measure and the triangle inequality we can estimate
$$vol(I)\leq vol(\bigcup_i I_i^*)\leq \sum_i vol(I_i^*),$$
and as the inequality holds for arbitrary $\{I_i^*\}_{i\in\mathbb{N}_1},$
we get that
$$vol(I)\leq \inf\left\{\sum_i vol(I_i^*):I\subset \bigcup_i I_i^*\right\}=\lambda^*(I).$$
To show that $\lambda^*(I)\leq vol(I)$, choose a collection $\{I_i^{**}\}_{i\in \mathbb{N}_1}$ such that $I_1^{**}=I$ and $I_{i\neq1}^{**}=\emptyset$. Now as
$$\sum_i vol(I_i^{**})\subset \left\{\sum_i vol(I_i^*):I\subset \bigcup_i I_i^*\right\},$$
by the definition of the infimum we get that
$$\lambda^*(I)=\inf\left\{\sum_i vol(I_i^*):I\subset \bigcup_i I_i^*\right\}\leq \sum_i vol(I_i^{**})=vol(I)+0=vol(I).$$
Now as $\lambda^*(I)\ge vol(I)$ and $\lambda^*(I)\leq vol(I)$, it has to hold that
$$\lambda^*(I)=vol(I).$$
What confuses me is that the professor of the course i'm studying this from spent over half an hour proving several lemmas for this result and also did the proof for open and closed $I$ separetly and i don't see why that would be necessary, is there some problem with the proof i proposed?
 A: Obviously $\lambda^*(I) \leq vol(I)$ since $I$ covers itself.
Your proof for $\lambda^*(I) \geq vol(I)$ is not convincing. You need to prove why $vol(I) \leq \sum_{i = 1}^{\infty}vol(I_i^*)$. For this you have to use the completeness of $\mathbb{R}^n$ because this statement is not true for $\mathbb{Q}^n$, since $\sum_{i = 1}^{\infty}vol(I_i^*)$ can be $0$.
Is $vol(\bigcup_{i = 1}^{\infty}I_i^*)$ even defined?, that is, is "geometric measure" even defined for sets that aren't $n$-intervals?
One way of proving this is to assume that $I$ is compact and therefore the problem reduces to the case of a finite cover, which is easier.
A: I think the usual approach here is to consider the case where $I$ is closed or not closed. In your proof of the second direction, you've allowed $I$ to be part of the covering set for $I$, but we do not know a priori that $I$ is closed (or open), for example $I$ could be $(0,1]$.
The usual definition of the Lebesgue outer measure requires covering by closed (or open) rectangles, and it is possible to prove that those two definitions are equivalent, but in general I don't think its common to define it in terms of rectangles that are neither open nor closed, which is why it might not be valid to include $I$ in your covering.
