Prove that a Lebesggue outer measure on $\mathbb{R}^d$ is an outer measure which assigns to each $d$-dimensional interval its volume.

Question

I just started to self-study some measure theory. I need to prove the following proposition

Proposition$\quad$ Lebesgue outer measure on $\mathbb{R}^d$ is an outer measure, and it assigns to each $d$-dimensional interval its volume.

where

Definition$\quad$ Let $X$ be a set, and let $\mathcal{P}(X)$ be the collection of all subsets of $X$. An outer measure on $X$ is a function $\mu^*:\mathcal{P}(X)\to[0,+\infty]$ such that

1. $\mu^*(\emptyset) = 0$,
2. if $A \subseteq B \subseteq X$, then $\mu^*(A)\leq\mu^*(B)$, and
3. if $\{A_n\}$ is an infinite sequence of subsets of $X$, then $\mu^*(\bigcup_nA_n)\leq\sum_n\mu^*(A_n)$.

and

Definition$\quad$ We define Lebesgue outer measure on $\mathbb{R}^d$ as follows. A $d$-dimensional interval is a subset of $\mathbb{R}^d$ of the form $I_1 \times \dots \times I_d$, where $I_1, \dots, I_d$ are subintervals of $\mathbb{R}$ and $I_1 \times \dots \times I_d$ is given by \begin{align*} I_1 \times \dots \times I_d = \{(x_1,\dots,x_d):x_i \in I_i\ \text{for } i=1,\dots,d\}. \end{align*} Note that the intervals $I_1, \dots, I_d$, and hence the $d$-dimensional interval $I_1 \times \dots \times I_d$, can be open, closed, or neither open nor closed. The volume of the $d$-dimensional interval $I_1 \times \dots \times I_d$ is the product of the lengths of the intervals $I_1, \dots, I_d$, and will be denoted by $\text{vol}(I_1 \times \dots \times I_d)$. For each subset $A$ of $\mathbb{R}^d$ let $\mathcal{C}(A)$ be the set of all sequences $\{R_i\}$ of bounded and open $d$-dimensional intervals for which $A \subseteq \bigcup_{i=1}^{\infty}R_i$. Then $\lambda^*(A)$, the Lebesgue outer measure of $A$ is the infimum of the set \begin{align*} \left\{\sum_{i=1}^{\infty}\text{vol}(R_i):\{R_i\}\in\mathcal{C}(A)\right\}. \end{align*}

Here is my attempt:

Proof$\quad$ We begin by verifying that $\lambda^*$ is an outer measure. The relation $\lambda^*(\emptyset)=0$ holds, since for each positive number $\epsilon$ there is a sequence $\{R_i\}$ of bounded and open $d$-dimensional intervals (whose union necessarily includes $\emptyset$) such that $\sum_i\text{vol}(R_i)<\epsilon$; for example, let $R_i = (a_1^i,b_1^i)\times\dots\times(a_d^i,b_d^i)$ where $b_1^i-a_1^i = \dots = b_d^i-a_d^i = \left(\frac{\epsilon}{3 \cdot 2^{i-1}}\right)^{\frac{1}{d}}$, so that $\text{vol}(R_i) = \frac{\epsilon}{3 \cdot 2^{i-1}}$, which implies that $\sum_i\text{vol}(R_i) = \frac{2\epsilon}{3} < \epsilon$.

For the monotonicity of $\lambda^*$, note that if $A \subseteq B$, then each sequence of bounded open $d$-dimensional intervals that covers $B$ also covers $A$, and so $\lambda^*(A) \leq \lambda^*(B)$.

Now consider that countable subadditivity of $\lambda^*$. Let $\{A_b\}_{n=1}^{\infty}$ be an arbitrary sequence of subsets of $\mathbb{R}^d$. If $\sum_n\lambda^*(A_n) = +\infty$, then $\lambda^*(\bigcup_nA_n) \leq \sum_n\lambda^*(A_n)$ certainly holds. So suppose that $\sum_n\lambda^*(A_n) < +\infty$, and let $\epsilon$ be an arbitrary positive number. For each $n$ choose a sequence $\{R_{n,i}\}_{i=1}^{\infty}$ of bounded open $d$-dimensional intervals that covers $A_n$ and satisfies \begin{align*} \sum_{i=1}^{\infty}\text{vol}(R_{n,i}) < \lambda^*(A_n) + \frac{\epsilon}{2^n}. \end{align*} If we combine these sequences into one sequence $\{R_j\}$, then the combined sequence satisfies \begin{align*} \bigcup_nA_n \subseteq \bigcup_jR_j \end{align*} and \begin{align*} \sum_j\text{vol}(R_j) < \sum_n\left(\lambda^*(A_n)+\frac{\epsilon}{2^n}\right) = \sum_n\lambda^*(A_n) + \epsilon. \end{align*} These relations, together with the fact that $\epsilon$ is arbitrary, imply that $\lambda^*(\bigcup_nA_n) \leq \sum_n\lambda^*(A_n)$. Thus, $\lambda^*$ is an outer measure.

Now we compute the outer measure of the $d$-dimensional intervals of $\mathbb{R}^d$. First consider a closed bounded, and thus compact, $d$-dimensional interval $K$. We cover $K$ with a sequence $\{R_i\}$ of bounded open $d$-dimensional intervals in which the first $d$-dimensional interval is barely larger than $K$, and the sum of the volumes of the other $d$-dimensional intervals is very small; for example, let $\epsilon$ be an arbitrary positive number and let $K \subseteq \bigcup_iR_i$ where $\text{vol}(R_1)=\text{vol}(K)+\epsilon$ and $\text{vol}(R_i)=\frac{\epsilon}{2^{i-1}}$ for $i>1$. Then, $\lambda^*(K) < \text{vol}(K)+2\epsilon$ for all $\epsilon>0$, and so $\lambda^*(K) \leq \text{vol}(K)$. We turn to the reverse inequality. Note that if $K$ is a compact $d$-dimensional interval and if $\{R_i\}_{i=1}^{\infty}$ is a sequence of bounded and open $d$-dimensional for which $K \subseteq \bigcup_{i=1}^{\infty}R_i$, then there is a positive integer $n$ such that $K \subseteq \bigcup_{i=1}^nR_i$, and $K$ can be decomposed into a finite collection $\{K_j\}_{j=1}^m$ that overlap only on their boundaries and are such that for each $j$ the interior of $K_j$ is included in some $R_i$ (where $i \leq n$). From this it follows that \begin{align*} \text{vol}(K) = \sum_{j=1}^m\text{vol}(K_j) \leq \sum_i\text{vol}(R_i) \end{align*} and hence that $\text{vol}(K) \leq \lambda^*(K)$. Therefore, $\text{vol}(K) = \lambda^*(K)$.

The outer measure of an arbitrary bounded $d$-dimensional interval is its volume, since such a $d$-dimensional interval includes and is included in closed bounded intervals of volume arbitrarily close to the volume of itself. Finally, an unbounded $d$-dimensional interval has infinite outer measure, since it includes arbitrarily larger closed bounded $d$-dimensional intervals.

Could someone please help me check if my proof is correct and rigorous? Thanks a lot in advance!

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To be more specific, I am not confident about the following parts of my proof:

1. When I prove $\lambda^*(\emptyset) = 0$, I constructed a sequences of bounded open $d$-dimensional intervals such that the series is less than $\epsilon$. However, I am not sure whether my construction is appropriate.
2. When proving the countable subadditivity under the assumption of $\sum_n\lambda^*(A_n)<+\infty$, without specifying what exactly the sequence $\{R_{n,i}\}$ is, I only expressed that I want $\sum_{i=1}^{\infty}\text{vol}(R_{n,i}) < \lambda^*(A_n) + \frac{\epsilon}{2^n}$ to be true. Is it rigorous enough to simply put it there? If not, how should I construct a sequence $\{R_i\}$ with its exact elements specified such that the inequality holds?
3. Similarly, when computing the outer measure of a compact $d$-dimensional interval, I failed to specify what exactly the covering sequence of bounded open $d$-dimensional intervals is; instead, I simply expressed that I want $\text{vol}(R_1)=\text{vol}(K)+\epsilon$ and $\text{vol}(R_i)=\frac{\epsilon}{2^{i-1}}$ for $i>1$ to hold. Is this correct and rigorous? If not, how should I improve it?
4. Finally, if it is necessary, how should one rigorously show that " $K$ can be decomposed into a finite collection $\{K_j\}_{j=1}^m$ that overlap only on their boundaries and are such that for each $j$ the interior of $K_j$ is included in some $R_i$ (where $i \leq n$)."