Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$R^2$, $A=\{(x,\;y)\in R^2\colon 0\leqslant x\leqslant 1,\;0\leqslant y\leqslant 1\}$. 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 $0$.

share|cite|improve this question
up vote 2 down vote accepted

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 $m_eX = 1$.

share|cite|improve this answer
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 $A$.

Suggestion: For a nice treatment of the Jordan measure (and measure theory in general), I strongly recommend Terence Tao's excellent book on Measure theory.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.