Union/intersection of two Jordan measurable sets is Jordan measurable using definition I know how to use $\partial(A \cup B) \subseteq \partial(A) \cup \partial(B)$ to prove the property for union but is it possible to prove the statement using the definition only?
 A: A set $A \subset \mathbb{R}^n$ is Jordan measurable if the inner and outer Jordan measures are equal, that is
$$m_*(A) :=\sup_{E \subset A} |E| = \inf_{E \supset A} |E| =: m^*(A) ,$$
where $E$ denotes an elementary set (finite union of non-overlapping rectangles) and $|E|$ is the volume of $E$. In this case we denote the Jordan measure  by $m(A) = m^*(A) = m_*(A)$. We have $m^*(A) > m_*(A)$ always if $A$ is not Jordan measurable.
It is easily shown that the Jordan measure of an elementary set is the volume (sum of volumes of composing rectangles), finite unions and intersections of elementary sets are elementary, and for elementary sets $E,F$ we have $|E\cup F|+ |E\cap F| = |E| + |F|$.
To prove that $A,B$ Jordan measureable implies that $A \cup B$ and $A\cap B$ are Jordan measurable, first consider that $A,B$ are bounded.
If $E,F$ are elementary sets such that $E \supset A$  and $F \supset B$, then 
$$m^*(A\cup B) + m^*(A\cap B) \leqslant |E\cup F| + |E\cap F| = |E| + |F|$$
Taking the infimum over all $E\supset A$ and $F\supset B$ it follows that
$$\tag{1}m^*(A\cup B) + m^*(A\cap B) \leqslant m^*(A) + m^*(B) = m(A) + m(B)$$
Similarly with elementary sets $E\subset A$ and $F \subset B$ it follows that
$$\tag{2}m_*(A\cup B) + m_*(A\cap B) \geqslant m_*(A) + m_*(B)= m(A) + m(B)$$
Subtracting (2) from (1) we get 
$$m^*(A\cup B) - m_*(A\cup B) + m^*(A\cap B)-m_*(A\cap B) \leqslant 0$$
Since, seperately, $m^*(A\cup B) - m_*(A\cup B) \geqslant 0$ and $m^*(A\cap B)-m_*(A\cap B) \geqslant 0$ and the sum of those differences is no bigger than $0$ we must have 
$$ m^*(A\cup B) = m_*(A\cup B), \quad m^*(A\cap B)=m_*(A\cap B), $$
and so the union and intersection are Jordan measurable.
Extension of the proof when at least one of $A$ and $B$ are unbounded is straightforward if that situation is allowed under your definition of Jordan measurability. 
