Proving equivalence of two definitions of Lebesgue measure. I got two definitions of Lebesgue measure of open sets. One is $$m_1(A)=\inf\Big\{\sum^{\infty}_{k=1}l(I_k):A\subset \bigcup_k I_k\Big\}$$ and the other is $$m_2(A)=\sup\Big\{\sum^{N}_{j=1} l(I_k):  \bigcup_k I_k \subset A \text{ &amp; } I_i^\circ \cap I_j^\circ= \emptyset \Big\}$$ where $I_k$ is rectangle in $\mathbb{R}^n$.


How can we show these two difinitions are equivalent? I want to prove this by writing $A$ as the union of countable rectangles with their interiors pairwise disjoint. But I'm not sure whether we can do this.

Any hints? Thanks in advance!
I believe it is possible, indeed here is my argument.

Let $A \in \mathbb{R}^n$ be open. 

Since A is open, for every $x \in A$ there is a ball $B_x$ with center $x$ that is inside $A$. Since the rationals $\mathbb{Q}^n$ are dense in $\mathbb{R}^n$, I can choose a (n-dimensional) rational $q_x$ and a radius $r_x$ such that the (n-dimensional) square $R_x$ with side $r_x$ and center $q_x$ is inside $B_x$ (and thus inside A) and $x \in R_x$. 

So by choosing for each $x$ one such rectangle $R_x$ I can define a function $F: x \mapsto R_x$ from $A$ to the set $S$ of all squares of rational center and rational radius. This set has a trivial bijection with $\mathbb{Q}^{n+1}$, so it is countable. $R := F(A)$ inside S is also countable, so let $R_n$ be an enumeration of this set. 

It is obvious that

$$
\bigcup_{n=0}^{\infty}{R_n} = A
$$

since every $R_n$ is inside A and all points of A are in some $R_n$. So we just wrote A as a countable union of rectangles.

We can now make this union with emtpy intersection of the interiors if we just define 

$$
B_n' := R_n \setminus \bigcup_{k=0}^{n-1}{R_k}
$$

and $B_n = \bar{B_n'}$. This works since the set that results from taking a finite number of rectangles from a rectangle can be written as a union of rectangles. We take care of boundaries by taking the closure.
I believe it is possible, indeed here is my argument.

Let $A \in \mathbb{R}^n$ be open. 

Since A is open, for every $x \in A$ there is a ball $B_x$ with center $x$ that is inside $A$. Since the rationals $\mathbb{Q}^n$ are dense in $\mathbb{R}^n$, I can choose a (n-dimensional) rational $q_x$ and a radius $r_x$ such that the (n-dimensional) square $R_x$ with side $r_x$ and center $q_x$ is inside $B_x$ (and thus inside A) and $x \in R_x$. 

So by choosing for each $x$ one such rectangle $R_x$ I can define a function $F: x \mapsto R_x$ from $A$ to the set $S$ of all squares of rational center and rational radius. This set has a trivial bijection with $\mathbb{Q}^{n+1}$, so it is countable. $R := F(A)$ inside S is also countable, so let $R_n$ be an enumeration of this set. 

It is obvious that

$$
\bigcup_{n=0}^{\infty}{R_n} = A
$$

since every $R_n$ is inside A and all points of A are in some $R_n$. So we just wrote A as a countable union of rectangles.

We can now make this union with emtpy intersection of the interiors if we just define 

$$
B_n' := R_n \setminus \bigcup_{k=0}^{n-1}{R_k}
$$

and $B_n = \bar{B_n'}$. This works since the set that results from taking a finite number of rectangles from a rectangle can be written as a union of rectangles. We take care of boundaries by taking the closure.