I have the following definition of area:
Let $A$ be a bounded set from $\mathbb{R}^2$. We say that $A$ has area if there exist two sequences $(E_n)_{n\in \mathbb{N}}, (F_n)_{n\in \mathbb{N}}$ of elementar sets, such that:
- (1) $E_n \subset A \subset F_n, \forall n \in \mathbb{N}$
- (2) the sequences of positive real numbers {$area(E_n)$}$_{n\in\mathbb{N}}$,{$area(F_n)$}$_{n\in\mathbb{N}}$ are convergent with $\lim_{n\rightarrow \infty}area(E_n)_{n\in\mathbb{N}}$=$\lim_{n\rightarrow \infty}area(F_n)_{n\in\mathbb{N}}$ (2).
In this case, we write:
$$area(A)=\lim_{n\rightarrow \infty}area(E_n)_{n\in\mathbb{N}}=\lim_{n\rightarrow \infty}area(F_n)_{n\in\mathbb{N}}$$
Now, a set $E\subset \mathbb{R}^2$ is called elementar if $E=\bigcup_{i=1}^nD_i$, where $D_i$ are rectangles with the sides parralel to the axes $(Ox, Oy)$ and any two different rectangles $D_i, D_j$ have at most one common side.
We define:
$$area(E):=\sum_{i=1}^n area(D_i)$$
The book where this definitions are presented omits to prove that the definition of $area(A)$ is correct, i.e. if we pick another two sequences $(U_n)_{n\in \mathbb{N}}$ and $(V_n)_{n\in \mathbb{N}}$ such that (1) and (2) are satisfied, then
$$\lim_{n\rightarrow \infty}area(E_n)_{n\in\mathbb{N}}=\lim_{n\rightarrow \infty}area(F_n)_{n\in\mathbb{N}}=\lim_{n\rightarrow\infty}area(U_n)_{n\in\mathbb{N}}$$
How can one prove this? Thank you very much!
More: if $A, B$ have area, then $A\cup B,$ A\B and $A\cap B$ have area.
Lets denote with $$a_n := area(E_n), \, b_n :=area(F_n), \, c_n=area(U_n), \, d_n=area(V_n)$$ and assume $\lim a_n = \lim b_n=:x$ and $\lim c_n = \lim d_n=:y$.
Your question is, why is $x = y $?
Because of $$E_n \subset A \subset V_n$$ we have $a_n \leq d_n$ and thus $x \leq y.$ Also we have $$U_n \subset A \subset F_n $$ which shows $c_n \leq b_n$ and thus $y \leq x$. All in all, we end up with $x=y$.