Completely stumped on exercise cooncerning the characterisation of Jordan measure - would anyone be willing to give a hint? In Terry Tao's notes on measure theory he has the following exercise, I have no idea how to deal with the last statement, I would really appreciate it if someone could give a hint for the final case.
Show that the following are equivalent:


*

*$E$ is Jordan Measurable

*For all $\epsilon > 0$ there exists elementary sets $A$ and $B$ satisfying $A \subset E \subset B$ such that $m(B/A) \leq \epsilon$

*For all $\epsilon >0$ there exists an elementary set $A$ such that $m^{*}(A\Delta E) \leq \epsilon,$ where $m^*(\cdot)$ is the Jordan outer measure.


Here is my attempt so far:
$1. \Rightarrow 2.$
Since $E$ is measureable, we have $m(E) = m^*(E) = m_*(E)$ where $m_*(\cdot)$ is the Jordan inner measure.
By assumption we can find elementary sets $B$ and $A$ where  $A \subset E \subset B$ such that $m(B) - m(E) \leq \epsilon$ and $m(E) - m(A) \leq \epsilon.$
It thus follows that $m(B) - m(A) \leq 2\epsilon.$
Since $A$ and $B$ are elementary we have that $m(B) = m(B/A) + m(A).$ combined with the inequality above this gives $m(B/A) \leq 2\epsilon$
$2. \Rightarrow 1.$
Choose $A \subset E \subset B$ with $m(B/A) \leq \epsilon$ then
$\epsilon \geq m(B/A) = m(B) - m(A) \geq m(B) - m_*(E) \geq m^*(E) - m_*(E) = |m^*(E) - m_*(E)|$
Since $\epsilon$ was arbitrary the result follows.
 A: $(2) \implies (3)$:
Let $\varepsilon>0$ be given, and let $A\subset E\subset B$ so that $A$ and $B$ are elementary and $m(B-A)<\varepsilon$. 
Since $A\subset E\subset B$ and by monotonicity, we have:
$$m^*(E\triangle B)=m^*(B-E)\le m^*(B-A)<\varepsilon.$$
For the implication $(3)\implies (1)$ we need the following lemma:
If $B$ is Jordan measurable and $E\subset B$ then
$$m^*(B-E)=m(B)-m_*(E).$$
Proof of the lemma:
Let $\varepsilon>0$ be given. Let $A$ and $C$ be elementery sets so that $A\subset E$, $B\subset C$, $m_*(E)-m(A)<\varepsilon$, and $m(C)-m^*(B)<\varepsilon$ (we have used the fact that $B$ is bounded). We have:
$$m^*(B-E)\le m^*(C-A)=m(C)-m(A)\le m^*(B)-m_*(E)+2\varepsilon.$$
Since $\varepsilon$ was arbitrary, we have:
$$m^*(B-E)\le m^*(B)-m_*(E)=m(B)-m_*(E).$$
For the reverse inequality, let $\varepsilon>0$ be given. Let $A$ and $C$ be elementery sets so that $A\subset B$, $(B-E)\subset C$, $m_*(B)-m(A)<\varepsilon$, and $m(C)-m^*(B-E)<\varepsilon$. Then $(A-C)\subset E$, so we have:
$$m(B)-m_*(E)=m_*(B)-m_*(E)$$ 
$$\le m(A)+\varepsilon-m(A-C)=m(C)+\varepsilon\le m^*(B-E)+2\varepsilon$$
Since $\varepsilon$ was arbitrary, we have:
$$m(B)-m_*(E)\le m^*(B-E).$$
Now, we are ready to prove that
$(3)\implies (1)$:
Let $\varepsilon>0$ be given, and let $E'$ be an elementery set so that $m^*(E\triangle E')<\varepsilon$. Let $B'$ be an elementary set so that $(E-E')\subset B'$ and $m(B')-m^*(E-E')<\varepsilon$. Since $(E-E')\subset (E\triangle E')$, it follows that $m(B')<2\varepsilon$.
Define $B=B'\cup E'$. Then $B$ is elementary and $E\subset B$. By the lemma, $m^*(B-E)=m(B)-m_*(E)$. Therefore:
$$m^*(E)-m_*(E)\le m(B)-m_*(E)=m^*(B-E)=m^*(E'\cup B'-E)$$
$$=m^*((E'-E)\cup (B'-E))\le m^*(E'-E)+m^*(B'-E)<\varepsilon + m(B')<3\varepsilon,$$
and we are done.
