Any succinct way to prove class of finite union of half-closed intervals form a ring of sets? I am learning measure theory and encountered two problems. I can prove the first one however very tedious through math induction. And I can not extend my method to $R^n$ since it is too complex. Does anybody have good way to prove it(Just prove $V'$ is closed under '-', i .e. $\forall A,B \in V'$, prove $A-B \in V'$  )?
1.V={ $\cup_{i=1}^{n}\ \left[a_i{,}\ b_i\right); a_i{,}\ b_i\in \ R\ {,}\ n\in \ N\ $  } $\cup${$\emptyset$}. Show $V$ is a ring of sets in $R$.
I used induction two times to prove this scenario, fist I show $\cup_{i=1}^{n}\ \left[a_i{,}\ b_i\right)-(a,b] \in V$(finite minus one), then I show $\cup_{i=1}^{n}\ \left[a_i{,}\ b_i\right)-\cup_{i=1}^{m}\ \left[c_i{,}\ d_i\right) \in V$.(finite minus finite) These two steps are both through induction.
Now consider $R^n$
2.V'={ $\cup_{i=1}^{m}\ \prod_{1}^{n}\left[a_i{,}\ b_i\right); a_i{,}\ b_i\in \ R\ {,}\ m\in \ N\ $  } $\cup${$\emptyset$}. Show $V'$ is a ring of sets in $R^n$.
I thought about this question for a while and now I know I only need to prove (one -one) is closed in $V'$. However how to prove it without discussing so many scenarios.
 A: Start with proving that set: $$\prod_{i=1}^n[a_i,b_i)-\prod_{i=1}^n[a'_i,b'_i)\tag1$$ can be written as a finite union of disjoint sets of the form $\prod_{i=1}^n[u_i,v_i)$. In this context we also accept the empty union (equalizing the empty set).
This shows that also: $$\prod_{i=1}^n[a_i,b_i)\cup\prod_{i=1}^n[a'_i,b'_i)=\left[\prod_{i=1}^n[a_i,b_i)-\prod_{i=1}^n[a'_i,b'_i)\right]\cup\prod_{i=1}^n[a'_i,b'_i)\tag2$$
can be written as finite union of disjoint set of that form.
Applying $(2)$ with repetition it can be shown that a union of the form: $$\bigcup_{k=1}^m\prod_{i=1}^n[a_{k,i},b_{k,i})$$can be written again as such union but then with the extra clause that the product sets are disjoint.
Accepting the empty set as the result of an empty union we can then write:$$V'=\left\{\text{ disjoint finite unions of sets of the form }\prod_{i=1}^n[a_i,b_i)\right\}$$
This makes it more easy to prove that $A,B\in V'$ implies that $A-B\in V'$.
Note that:$$\bigcup_{k=1}^m\prod_{i=1}^n[a_{k,i},b_{k,i})-\bigcup_{r=1}^s\prod_{i=1}^n[a'_{r,i},b'_{r,i})=\bigcup_{k=1}^m\left[\prod_{i=1}^n[a_{k,i},b_{k,i})-\bigcup_{r=1}^s\prod_{i=1}^n[a'_{r,i},b'_{r,i})\right]$$and applying repetition we arrive at a disjoint finite union of product sets.

Addendum
Let us call a collection $\mathcal A\subseteq\mathcal P(X)$ good if it is closed under binary intersections and for every pair $A,A'$ the set $A-A'$ can be written as a finite union of disjoint elements of $\mathcal A$.
It is our aim to prove that for good collections $\mathcal A\subseteq\mathcal P(X)$ and $\mathcal B\subseteq\mathcal P(Y)$ also the collection:$$\mathcal C=\{A\times B\mid A\in\mathcal A, B\in\mathcal B\}\subseteq\mathcal P(X\times Y)$$is good.
It is straightforward that $\mathcal C$ is closed under binary intersections so it is enough to prove that for $A,A'\in\mathcal A$ and $B,B'\in\mathcal B$ the set $(A\times B)-(A'\times B')$ can be written as finite union of disjoint elements of $\mathcal C$.
Note that $D:=A\cap A'\in\mathcal A$ and we can write the set as disjoint union:$$(A\times B)-(A'\times B')=[(A-A')\times B]\cup[D\times(B-B')]]$$
Here $A-A'$ and $B-B'$ can both be written as finite union of disjoint elements in $\mathcal A$ and $\mathcal B$ respectively and this leads evidently to a finite union of disjoint elements in $\mathcal C$ on RHS.
So we are ready.
Applying this to the situation where $X=Y=\mathbb R$ and $\mathcal A=\mathcal B=\{[a,b)\mid a<b\}\subseteq\mathcal P(\mathbb R)$ we find that for $n=2$ the collection $\left\{\prod_{i=1}^n[a_i,b_i)\mid a_i<b_i\text{ for }i=1,\dots,n\right\}\subseteq\mathcal P(\mathbb R^n)$ is good and by induction on $n$ this can be proved for every $n$.
