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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 $E$.

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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.

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