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By definition, Jordan outer measure of a subset $E$ in $\mathbb{R}^n$ is the approximation to area of $E$ by finitely many open cubes(rectangles) which cover $E$. Similarly, the Jordan inner measure is the approximation to the area of $E$ by finitely many closed cubes (rectangles) which lie in $E$.

$E$ is said to be Jordan measurable if $E$ has same inner and outer Jordan measure.

Question: If $E_1\subseteq E_2\subseteq \cdots$ is an ascending sequence if Jordan measurable sets, and if $E=\cup_n E_n$ is Jordan measurable, is it true that the limit of Jordan measures of $E_n$ is the Jordan measure of $E$?

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  • $\begingroup$ It's easy to see that this is true if you've already shown that Lebesgue measure extends Jordan measure. If you're trying to avoid that machinery for the proof, though, things might be a bit trickier, since Jordan measure won't play as nicely with limits... $\endgroup$ – Josh Keneda Sep 20 '14 at 4:39
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One must show that $\,m(E)=\sup \,\{m(E_n)\}\,$.

Let $E_0=\emptyset\,$ and $\,A_n=E_n\setminus E_{n-1}\,$.

For every $\varepsilon>0\,$ choose a finite union of closed cubes $\,F\subset E\,$ such that $$m(F)>m(E)-\varepsilon$$ and finite unions of open cubes $\,G_n\supset A_n\,$ such that$$m(G_n)<m(A_n)+ \frac \varepsilon {2^n}$$ By compactness of $F$, there exists $N$ such that $$F \subset G_1 \cup \dots \cup G_N$$ Hence $$m(E)- \varepsilon< m(A_1)+ \dots m(A_N)+\varepsilon=m(E_N)+\varepsilon$$ It follows that $$m(E_N)>m(E)-2\varepsilon$$

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