what ensures the existence of a closed set in a theorem about measure of elementary sets I am going through the text by Kolmogrov and Formin  on Lebesgue integration . So the book starts with laying down the foundation with elementary sets and defining a notion of measure on them .
The definition of an elementary set is a set in the X-Y Plane which can be decomposed into a finite combination of disjoint rectangles .
So , the second theorem that it presents goes as follows :
If $\ A $ is an elementary set which is contained in finite or countable system of elementary sets $\ A_n$ , then  $\ m(A) $  < $\ m( \cup A_n) $
The theorem starts by asserting that inside the set $\ A $ , one can find a closed set , $\ C $ such that $\ m(A) - m(C) < \epsilon/2 $ .
Is there any rigorous argument for this which does not take into account the fact that $\ A $ is an elementary set , i.e. , the specificity of it being a subset of $\ R^2 $ ? Can we provide a different argument which does not rely on the fact that this elementary set is a finite combination of rectangles ? I realize the possibility of such an argument relying on tools being developed in the following parts of the book when it generalizes beyond rectangles .
 A: My guess is they take $C$ to be a union of closed rectangles contained in the rectangles comprising $A$; it wouldn't make sense to me otherwise, since I'm guessing so far they've only defined the measure $m$ for such elementary sets.
You're right that this can be done in general (for Lebesgue measure), and it's called inner regularity (there they use compact sets, but for elementary sets they are the same).
As for how you'd show such a thing for a general set $A$; in some sense it's almost by definition. Recall (or you'll see later in the chapter) that we can define the Lebesgue measure as (don't be alarmed if it's slightly different in your source, but try to prove that they're equivalent!)
$$
m^*(A)=\inf\left\{ \sum_n m(A_n): A\subset \cup_n A_n, \, A_n \text{ open rectangle}\right\}.
$$
This extends $m$ from elementary sets to the whole collection of subsets of $\mathbb{R}^2$. Notice from this definition that it's immediate that we can approximate the measure of a set $A$ by that of an open set containing it.
Well in a similar (but not the same!) spirit, we can define another extension of $m$ by looking at approximations from below:
$$
m_*(A)= \sup\left\{ m(C): C\subset A,\, C\text{ closed/compact}\right\}.
$$
The crux of the matter is proving that these two extensions agree on, say, bounded Lebesgue measurable sets (in fact it may be an exercise later on in your book). Then the property that you want follows from this.
NOTE: $m_*$ needs that we already know $m$ on Lebesgue sets; indeed it'd be pretty useless if we took the supremum over elementary sets $C$ since it'd assign measure zero to pretty much every set with empty interior.
