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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.

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2 Answers 2

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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.

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  • $\begingroup$ Good explanation. To handle non-open $E,$ the definition of $\partial E$ should be $\overline{E}\setminus\operatorname{int}(E).$ $\endgroup$
    – Dap
    Feb 2, 2018 at 12:02
  • $\begingroup$ @Dap: thanks for pointing that out. Somehow I had missed that part. I have fixed my post now. $\endgroup$
    – Paramanand Singh ♦
    Feb 2, 2018 at 13:14
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Using the Heine-Borel property you can prove that compact sets have zero Jordan measure if and only if they have zero Lebesgue measure.

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  • $\begingroup$ 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 ? $\endgroup$
    – Our
    Feb 2, 2018 at 10:18
  • $\begingroup$ @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 $\endgroup$
    – Dap
    Feb 2, 2018 at 10:46
  • $\begingroup$ 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 ? $\endgroup$
    – Our
    Feb 2, 2018 at 10:51
  • $\begingroup$ In other words, "So you are saying" was denoting that "So you are leading me to think as that" $\endgroup$
    – Our
    Feb 2, 2018 at 10:52
  • $\begingroup$ 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. $\endgroup$
    – Our
    Feb 2, 2018 at 10:59

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