Finite additivity of Jordan measure Let $E, F\subset \mathbb{R^d}$ be Jordan measurable sets. I have to show that $E  \cup F$ is Jordan measurable, that is, $sup_{A \subset E \cup F,\space A\space elementary}m(A) = inf_{E \cup F \subset B,\space B\space elementary}m(B)$, where m(A) is the elementary measure of an elementary set A. (Exercise 1.1.6 of Terence Tao's An Introduction to Measure Theory)
E,F are Jordan measurable iff for every $\epsilon > 0$, there are sets $A_{1} \subset E \subset B_{1}$, $A_{2} \subset F \subset B_{2}$ such that m($B_{1}$ \ $A_{1}$) < $\epsilon$ and m($B_{2}$ \ $A_{2}$) < $\epsilon$.  
 A: Hints: Let $\epsilon > 0$. Using your definitions, there are elementary sets $A_1 ⊂ E ⊂ B_1$ and $A_2 ⊂ E ⊂ B_2$ with $m(B_i\setminus A_i) <\epsilon/2$ for $i = 1, 2.$ 
Now note that $A_1 ∪ A_2 ⊂ E ∪F ⊂ B_1 ∪ B_2$ and $(B_1 ∪ B_2) \setminus (A_1 ∪ A_2) ⊂ (B_1 \setminus A_1)∪(B_2 \setminus A_2)$
A: First, we demonstrate that $E \cup F$ is Jordan-measurable, i.e., \begin{equation*}\sup \left\{ m(A) : \begin{array}{l} A\subset (E \cup F) \\ \text{$A$ is elementary}\end{array} \right\} = \inf \left\{ m(B) : \begin{array}{l} (E \cup F) \subset B \\ \text{$B$ is elementary}\end{array} \right\} \end{equation*} As always, we show that we can always find an outer cover $\overline{E \cup F}$ and an inner cover $\underline{E \cup F}$ such that the difference between their simple measures does not exceed $\epsilon>0$. Thus, fix an arbitrary $\epsilon$. Since $E,F$ are Jordan-measurable, we can find outer covers $\overline{E},\overline{F}$ and inner covers $\underline{E},\underline{F}$ such that $m\left(\overline{E}\right) - m\left(\underline{E}\right) \leq \epsilon/2$ and $m\left(\overline{F}\right) - m\left(\underline{F}\right) \leq \epsilon/2$ by Exercise 1.1.5. Define \begin{equation*} \overline{E \cup F}:=\overline{E} \cup \overline{F} \hspace{1cm} \underline{E \cup F}:= \underline{E} \cup \underline{F} \end{equation*} The properties of the elementary measure ensure that the $\epsilon/2$ covering we have chosen gives us the desired result: \begin{equation*} \begin{array}{ll} m\left(\overline{E \cup F}\right) - m\left(\underline{E \cup F}\right) &= \left( m\left(\overline{E}\right) + m\left(\overline{F}/\overline{E}\right) \right) - \left( m\left(\underline{E}\right) + m\left(\underline{F}/\underline{E}\right) \right)\\[5pt] &= \left( m\left(\overline{E}\right) - m\left(\underline{E}\right) \right) + \left(m\left(\overline{F}/\overline{E}\right) - m\left(\underline{F}/\underline{E}\right) \right)\\[5pt] &\leq \left( m\left(\overline{E}\right) - m\left(\underline{E}\right) \right) + \left(m\left(\overline{F}\right) - m\left(\underline{F}\right) \right)\\[5pt] &\leq \epsilon/2 + \epsilon/2 = \epsilon \end{array} \end{equation*}
This proof is taken from the solution manual to Terence Tao's An Introduction to Measure Theory.
