# Equivalent definition of Jordan measure

A set $$E$$ is Jordan-measurable if its outer Jordan measure and its inner Jordan measure coincide.

I am trying to prove that the following is an equivalent definition

A set $$E$$ is Jordan measurable if for every $$\epsilon>0$$, there exists an elementary set $$A$$ such that $$m^*(A \triangle E) < \epsilon$$, where $$m^*$$ denotes the Jordan outer measure.

I got stuck here because although $$A$$ is elementary, $$A \cap E$$ is not necessary an elementary set. So I do not see how to find a proper way to lower bound the inner Jordan measure of $$E$$.

Suppose $$E$$ is Jordan measurable. Then, for every $$\varepsilon>0$$, there exist two elementary sets $$A,B$$ such that $$A\subset E\subset B$$ and $$m(B\setminus A)<\varepsilon$$. So you obtain the condition because $$A\triangle E \subset B\setminus A$$.
Conversely, suppose the condition is valid. Then there exists an elementary set $$A'$$ such that $$A\triangle E \subset A'$$ and $$m(A')<\varepsilon$$.
The sets $$A\setminus A'$$ and $$A\cup A'$$ are elementary. Moreover you have $$A\setminus A'\subset E\subset A\cup A'$$ and $$A\cup A'\setminus (A\setminus A')=A'.$$
That is enough to say that $$E$$ is Jordan measurable.