Show that if $E$ is Jordan measurable then $m(A-B) \leq \epsilon$ I want to show that if $E$ is Jordan measurable then for every $\epsilon > 0$ there exists $A \subset E \subset B$ such that $m(A-B) \leq \epsilon$.
I think I have the right ideas but feel I am missing some details. I'd like some feedback for the following proof:
Let $\epsilon>0$ be given. Suppose $E$ is Jordan measurable. Then there exists $A$ and $B$ elementary with $A \subset E \subset B$ such that
$$
\sup m(A) = \inf m(B)
$$
and thus
$$
m(A) \leq m(B)
$$
which implies 
$$
0 \leq m(B) - m(A)
$$
Now since $A \subset B$ we know
$$
0 \leq m(B) - m(A) = m(B-A)
$$
and we can always choose $A,B$ such that
$$
m(B-A) \leq \epsilon
$$ 

There's maybe a typo in the problem statement? I'm assuming that what you mean is "if $E$ is Jordan measurable, then for every $\epsilon > 0$ there exist elementary sets $A \subseteq E \subseteq B$ such that $m(B-A) \leq \epsilon$" -- is that right?
Assuming so --
You say that because $E$ is measurable, there exist $A \subseteq E \subseteq B$ elementary such that $\sup m(A) = \inf m(B)$. I'm not sure what that means. Since $A$ is elementary, $m(A)$ is a number, not something you can take sup or inf of.
From the definition of Jordan measurable, since $E$ is measurable you have that $\sup_{A \subseteq E} m(A) = \inf_{E \subseteq B} m(B) = m(E)$ where the inf and sup are taken over all elementary subsets (resp. supersets).
Since the inf and sup are taken over all such subsets, you can now use properties of inf and sup to choose $A$ and $B$ such that the differences $m(E) - m(A)$ and $m(B) - m(E)$ are small.
EDIT: okay, so once you have $\sup_{A \subseteq E} m(A) = \inf_{E \subseteq B} m(B) = m(E)$, which is just the definition of Jordan measurable, what you want to do is take some $\epsilon > 0$. Use definition of sup/inf to choose $A \subseteq E$ elementary such that $m(A) > m(E) - \epsilon/2$ and $B \supseteq E$ elementary such that $m(B) < m(E) + \epsilon/2$. combining these yields $m(B) - m(A) < \epsilon$. Now use additivity of measure over disjoint unions to get $m(B-A) = m(B) - m(A)$.
