In the book of Mathematical Anaylsis II by Zorich, at page 122, it is given that
Definition: A set E is Jordan-measurable if it is bounded and its
boundary has Jordan measure zero.
Remark:
As Remark 2 shows, the class of Jordan-measurable subsets is precisely
the class of admissible sets introduced in Definition 1.
However, in order for Jordan measure to be defined on a set $E$, we only need its boundary to be Lebesgue measure zero, not Jordan measure zero, so I'm little confused about the definition.
To clarify, if a set is admissible, by definition of Jordan measure, we can define the Jordan measure of that set. Similarly, if $E$ is bounded and its boundary is Lebesgue measure zero, by definition, $E$ is admissible, so I do not understand behind the motivation for using Jordan measure zero instead of Lebesgue measure zero in the definition of Jordan measurable sets.
Best Answer
First we discuss admissible sets. From your post it appears that a set $E\subseteq \mathbb{R} ^{n} $ is admissible if $E$ is bounded and its boundary $\partial E$ is of Lebesgue measure $0$. Next note that $\partial E=\bar{E} - \operatorname {int} E=\bar{E} \cap (\operatorname {int} E) ^{c} $ and thus $\partial E$ is the intersection of two closed sets and is therefore closed itself. It follows that $\partial E$ is closed as well as bounded and hence compact (we are dealing in $\mathbb{R} ^{n}$). Since it is of Lebesgue measure $0$ therefore given any $\epsilon>0$ it has a countable open cover of total length less than $\epsilon$. By compactness we can find a finite subcover for $\partial E$ and afortiori its length is less than $\epsilon$. It thus follows that Jordan measure of $\partial E$ is $0$. Thereby according to the definition in your post $E$ is Jordan measurable.
The other implication is trivial. If we have a set $E$ which is Jordan measurable then by definition $\partial E$ has Jordan measure $0$ and therefore for any $\epsilon >0$ there is a finite cover for $\partial E$ which is of length less than $\epsilon$. Since a finite cover is also countable, it follows that Lebesgue measure of $\partial E$ is $0$ and thus by definition $E$ is an admissible set.