# Why Jordan measure is undefined?

$R^2$, $A=\{(x,\;y)\in R^2\colon 0\leqslant x\leqslant 1,\;0\leqslant y\leqslant 1\}$. Consider $X=A\cap Q^2$. Why for $X$, $m_e X=1,\;m_i X=0,\;m_e X\neq m_i X$? Especially i interested in why inner measure equals to $0$.

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by definition we have $m_iX = \sup_{S \subseteq X \text{ simple}} m(S)$ where a set is simple iff it is the finite disjoint union of sets of the form $[a,b) \times [c,d)$. Since a notempty half-open interval has inner points, every non-empty simple set has also. As $X$ doesn't have an inner point, $m_iX = 0$.

For $m_eX = 1$, as $m_eX = \inf_{S \supseteq X \text{ simple}} m(S)$ we obviously have $m_eX \le 1$ as $[0, 1+\epsilon)^2$ is a simple set of measure $(1+\epsilon)^2$ which contains $X$. To show $m_eX \ge 1$ let $S \supset X$ be simple. We show $[0,1)^2 \subseteq S$, as $m$ is monotone on simple sets, the conclusion follows. So let $(x,y) \in [0,1)^2$, if we had $(x,y) \not\in S$, there were an $\epsilon > 0$ such that $[x, x+\epsilon) \times [y,y+\epsilon)$ is disjoint from $S$ (as $S$ is a finite union of such intervals), which contradicts $A \subseteq S$. So $m_eX \ge 1$ and we get $m_eX = 1$.

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sorry, but why $X$ doesnt have inner points? its in $Q^2$. And $Q$ is subset of $R$. –  Yola May 15 '12 at 13:53
An inner point of $A$ were a point $(x,y) \in A$ with $(x-\delta, x+\delta) \times (y-\delta, y+\delta) \subseteq A$ for some $\delta > 0$. $A$ doesn't have inner points as $\mathbb Q \subseteq \mathbb R$ doesn't. –  martini May 15 '12 at 13:57
Hint: One can show that the Jordan outer measure of a bounded set $E$ is always equal to the Jordan outer measure of the closure of $E$. You can also show that the Jordan inner measure is bounded above by the Lebesgue outer measure of a bounded set $E$. Now it is not that difficult to show that the Jordan outer measure and the Lebesgue outer measure disagree on $A$.