# A confusion about the definition of Jordan measurable set

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.

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.

• Good explanation. To handle non-open $E,$ the definition of $\partial E$ should be $\overline{E}\setminus\operatorname{int}(E).$
– Dap
Feb 2, 2018 at 12:02
• @Dap: thanks for pointing that out. Somehow I had missed that part. I have fixed my post now. Feb 2, 2018 at 13:14

Using the Heine-Borel property you can prove that compact sets have zero Jordan measure if and only if they have zero Lebesgue measure.

• So you are saying that for any $E$, consider $\bar E$, since $\bar E$ is compact, for $E \subset \bar E$, using Jordan measure zero or Lebesgue measure zero is the same thing because we always have $E \subset \bar E$, right ?
– Our
Feb 2, 2018 at 10:18
• @onurcanbektas: no, I am just saying compact sets (such as the boundary of a bounded set) have Jordan measure zero if and only if they have Lebesgue measure zero
– Dap
Feb 2, 2018 at 10:46
• Ok, but that argument is not complete for the my question because $E$ does not have to be closed, and I'm trying to compete it to a full answer, and now asking that whether my argument is correct ?
– Our
Feb 2, 2018 at 10:51
• In other words, "So you are saying" was denoting that "So you are leading me to think as that"
– Our
Feb 2, 2018 at 10:52
• I missed the point, you are right. The boundary is always compact, hence there is no difference between being Lebesgue measure zero and being Jordan measure zero.
– Our
Feb 2, 2018 at 10:59