Exercise on Boolean closure of elementary sets I am reading Tao's notes on measure theory and I am stuck with an exercise, so here is the problem:
Definition We define $|I|$ the lenght of an interval of endpoints $a<b$ to be $|I|=b-a$. A box in $\mathbb R^d$ is a cartesian product $B=I_1 \times...\times I_d$ of $d$ intervals. An elementary set is any subset of $\mathbb R^d$ which is the union of a fnite number of boxes.
Show that if $E,F \subset \mathbb R^d$ are elementary sets, then $E \cup F$, $E \cap F$, $E \setminus F$ and $E \Delta F$ are elementary sets. If $x \in \mathbb R^d$ show that the translate $E+x=\{y+x : y \in E\}$ is an elementary set.
Attempt at a solution
If $E$ and $F$ are elementary sets then $E=B_1 \cup ... \cup B_k$ and $F=S_1 \cup ... \cup S_n$ union of $k$ and $n$ boxes respectively , so $E \cup F=(B_1 \cup ... \cup B_k) \cup (S_1 \cup ... \cup S_n)$ the union of $k+n$ boxes, i.e., $E \cup F$ is an elementary set by definition. 
I don't know how to show that the intersection and $E \setminus F$ are elementary. If I could prove these, then $E \Delta F= (E \setminus F) \cup (F \setminus E)$ is union of elementary sets which is an elementary set. 
For the translation, I know that $E=B_1 \cup ... \cup B_k$, so I can write $E+x=B_1+x \cup ... \cup B_k+x$. For each $B_i, 1\leq i \leq k$, $B_i=I_1 \times ... \times I_n$, so $B_i+x=I_1+x \times ... \times I_n+x$ and if an interval $I$ has endpoints $a<b$, then $I+x$ is an interval of endpoints $a+x<b+x$, so $E+x$ is a finite union of boxes, i.e., $E+x$ is an elementary set.
I would appreciate some help with the two remaining proofs.
 A: For the intersection, this is what I have.
$E \bigcap F = \left(\bigcup\limits_{i=1}^mA_i \right) \bigcap \left(\bigcup\limits_{j=1}^nB_j \right)$ = $\bigcup_{{1\leq i \leq m} \quad {1\leq j \leq n}} \left( A_i \bigcap B_j \right).$ We turned the intersection of unions into a union of intersections.
A: For the set theoretic difference $E\setminus F$:
Let $E=\cup_{i=1}^n A_i$ and $F=\cup_{j=1}^m B_j$ where $\lbrace A_i\rbrace_{i=1}^n$ and $\lbrace B_j\rbrace_{j=1}^m$ are sets of boxes. Note that if $A$ and $B$ are boxes, then $A\setminus B$ is elementary. Now, for each $j$ we have $E \setminus B_j = (\cup_{i=1}^n A_i)\setminus B_j = \cup_{i=1}^n (A_i \setminus B_j)$. Since each $A_i$ and $B_j$ is a box, each $A_i\setminus B_j$ is elementary, so $E \setminus B_j$ is a finite union of elementary sets. It then follows from the result on unions, that $E \setminus B_j$ is elementary. Since $E \setminus F = E \setminus (\cup_{j=1}^m B_j) = \cap_{j=1}^m (E\setminus B_j)$, we see that $E\setminus F$ is a finite intersection of elementary sets, and is therefore elementary by the result for intersections.
