Difference of Jordan Mesurable Sets is Measurable I have to prove that the set difference of Jordan Measurable Sets is Measurable; I am wondering if I can prove it by using the definition of measurable sets (i.e. upper and lower coincide)? I know how to prove it using $A_1 ⊂ E ⊂ B1$ and $A_2 ⊂ E ⊂ B_2$ with $m(B_i\backslash A_i) <\epsilon / 2$
for $i = 1, 2.$
 A: We have the definition that $A \subset \mathbb{R}^n$ 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 $E$. When $A$ is Jordan measurable, we denote the Jordan measure  by $m(A) = m^*(A) = m_*(A)$. 
Suppose that $A \subset B$ where $A,B$ are bounded and Jordan measurable. To prove that $B \setminus A$ is Jordan measureable we must show that $m_*( B \setminus A) = m^*(B \setminus A)$.
Take any rectangle $Q$ where $A \subset B \subset Q$ and note that
$$Q \setminus(B\setminus A) = Q \cap(B \cap A^c)^c = (Q \cap B^c)\cup (Q \cap A) = (Q\setminus B) \cup A $$
We then have
$$\tag{1}m^*(Q \setminus(B\setminus A)) = m^*((Q\setminus B) \cup A ) \leqslant m^*(Q\setminus B) + m^*(A),$$
where the last inequality is obtained form the property of subbaditivity of Jordan outer measure -- $m^*(A_1 \cup A_2) \leqslant m^*(A_1) + m^*(A_2)$ -- which I leave as an exercise to prove.
Now we need another basic property that for any subset $C$ of a rectangle $Q$ we have
$$\tag{2} m^*(Q \setminus C) = |Q| - m_*(C)$$
Deferring the proof of (2) and applying it to (1) first with $C = B\setminus A$ on the LHS and then with $C = B$ on the RHS we get,
$$|Q| - m_*(B \setminus A) \leqslant |Q| - m_*(B) + m^*(A),$$
which implies
$$\tag{3} m_*(B \setminus A) \geqslant m_*(B) - m^*(A)$$
By a similar argument we can prove that 
$$\tag{4} m^*(B \setminus A) \leqslant m^*(B) - m_*(A)$$
With (3) and (4) we have proved the intuitively obvious facts that the outer (inner) measure of a set difference is no bigger (smaller) than the outer (inner) measure of the superset minus the inner (outer) measure of the subset.
Substracting (4) from (3) we get
$$ 0 \leqslant m^*(B \setminus A) -  m_*(B \setminus A) \leqslant m^*(B) - m_*(B ) +   m^*(A) - m_*(A) = 0,$$
since $A$ and $B$ are by hypothesis Jordan measurable with equal inner and outer measures.
Therefore, $m^*(B \setminus A) = m_*(B \setminus A)$ and $B \setminus A$ is Jordan measureable. 
