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$\bar{E}$ is the closure of E.

How to prove $E\subset \mathbb R^n$ is Jordan-measurable is equivalent to $\bar{E}-E$ has Jordan measure null.

Definition of Jordan Measure: http://en.wikipedia.org/wiki/Jordan_measure.

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  • $\begingroup$ What is your definition of Jordan-measurable? $\endgroup$
    – leo
    Nov 7 '11 at 8:30
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    $\begingroup$ I guess you want $E$ to be bounded... $\endgroup$ Nov 7 '11 at 16:28
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Take $R$ a closed bounded interval (box) such that $E\subseteq R$

First suppose that $E$ is Jordan measurable. Let $\epsilon\gt 0$. Then exist $I_1,\ldots I_p$ and $J_1,\ldots J_q$ closed bounded intervals such that $\bigcup_1^p I_k\subseteq E\subseteq \bigcup_1^p J_k$, and $$\begin{align*} V(E)-\frac{\epsilon}{2}&\lt \sum_{k=1}^p I_k\\ \sum_{k=1}^q J_k &\lt V(E)+\frac{\epsilon}{2}. \end{align*}$$ Take $P=\{R_1,\ldots,R_m,R_{m+1},\ldots,R_{m+l},\ldots,R_s\}$ a partition of $R$ such that the intervals $I_k$ and $J_k$ are unions of intervals of $P$. The intervals of $P$ are numbered in the following way: $$\begin{align*} \bigcup_{k=1}^m R_k &= \bigcup_{k=1}^p I_k\\ \bigcup_{k=1}^l R_{m+k} &\subseteq \bigcup_{k=1}^q J_k, \end{align*}$$ I mean, the $R_{m+1},\ldots,R_{m+l}$ are the remaining intervals of $P$ needed to get: $$\bigcup_{k=1}^{m+l} R_k=\bigcup_{k=1}^q J_k.$$

Then $$\bar{E}-E\subseteq \bigcup_{k=1}^l R_{m+k}.$$ To prove the last affirmation, take $x\in\bar{E}-E$. Then, there is a sequence $(x_n)$ in $E$ such that $x_n\to x$. Since $E\subseteq \bigcup_1^q J_k$, there exist a subsequence $(x_{n_j})$ tottally contained in some $J_r$ (pinge-onhole principle) such that $x_{n_j}\to x$. Since $J_r$ is closed, $x\in J_r$. Now, note that since $x\not\in E $, $J_r\not\subseteq E$, and then by the choose of the $R_{m+1},\ldots,R_{m+l}$, $J_r$ is the union of some of this rectangles and then $x\in\bigcup_{k=1}^l R_{m+k}$. Then $$\sum_{k=1}^l V(R_{m+k})=\sum_{k=1}^q V(J_k) -\sum_{k=1}^p V(I_k) \lt \left(V(E)+\frac{\epsilon}{2}\right)-\left(V(E)-\frac{\epsilon}{2}\right).$$

To prove the converse, first an observation: let $A$ any subset of $R$, let $P=\{R_1,\ldots,R_s\}$ a partition of $R$. Consider $\Delta=\{k\in\{1,\ldots,s\}:R_k\cap A\neq\emptyset\}.$ Then $A\subseteq \bigcup_{k\in\Delta} R_k$. Note that is enough to show that given $\epsilon\gt 0$ we can find an approximation by excess and other by defect such that their difference is less than epsilon. Now, let $\epsilon\gt 0$. Take intervals $\{R_k\}_{k=1}^p$ such that $\bar{E}-E\subseteq \bigcup_1^p R_k$ and $\sum_{k=1}^pR_k\lt \epsilon$. We can assume that $\bar{E}\subseteq R$. Take a partition $P=\{P_1,\ldots,P_m\}$ of $R$ such that the $R_k'$s are unions of $P_k'$s. We can divide the set $\{1,\ldots,m\}$ in 3 disjoint sets $\Delta_1=\{k\in\{1,\ldots,m\}:P_k\subseteq E\}$, $\Delta_2=\{k\in \{1,\ldots,m\}:P_k\cap (\bar{E}-E)\neq \emptyset\}$, $\Delta_3=\{k\in\{1,\ldots,m\}:P_k\cap (\bar{E}-E)=\emptyset\}$. The last step is to show that $E\subseteq \bigcup_{k\in\Delta_2} P_k$, $\bigcup_{k\in\Delta_1} P_k\subseteq E$ and $\sum_{k\in\Delta_2}V(P_k)-\sum_{k\in\Delta_1} V(P_k)\lt\epsilon.$

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  • $\begingroup$ I'm sorry I think there is something wrong in your proof. In fact this is not a crect proposition. Consider a fat Cantor set $F$.$F$ is not Jordan measurable,but $\bar{F}-F=\emptyset$ is null. $\endgroup$
    – Leitingok
    Nov 18 '11 at 14:44
  • $\begingroup$ @Leitingok: Thanks for pointing that out. But your example shows that $V(\bar{X}-X)=0$ does not implies that $X$ is Jordan measurable. So, $X$ Jordan measurable is not equivalent to $\bar{X}-X$ measurable with $V(\bar{X}-X)=0$. Why are you asking for a "not correct proposition"? $\endgroup$
    – leo
    Nov 18 '11 at 19:34
  • $\begingroup$ But, certainly if $X$ is Jordan measurable then $V(\bar{X}-X)$ is measurable with $V(\bar{X}-X)=0$. $\endgroup$
    – leo
    Nov 18 '11 at 19:37

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