I cannot understand:

$$\bar{\mu} ( \{Q \cap (0,1) \} ) = 1$$

and (cannot understand this one particularly)

$$\underline{\mu} ( \{Q \cap (0,1) \} ) = 0$$

where $Q$ is rational numbers, why? I know that the measure for closed set $\mu ([0,1]) = 1$ so I am puzzled with the open set solution. Is $\underline{\mu} ( (0,1) ) = 0$ also? How is the measure with open sets in general?

So far the main question, history contains some helper questions but I think this is the lion part of it what I cannot understand. More about Jordan measure here.


  1. Jordan measure with semi-closed sets here
  2. Jordan measure and uncountable sets here

The inner measure of $S:=\mathbb Q \cap [0,1]$ is, by definition, the "largest" simple subset contained in S (, with largest defined as the sup of the measures of simple sets contained in $S$). But there is no non-trivial , i.e., non-empty, simple set contained in $S'$, since , by density of both rationals and irrationals in $\mathbb R$, any simple set $S':=[a,a+e)$ contained in $S$ (i.e., with $0<a<a+e<1$ , will necessarily hit some irrational, i.e., no non-empty simple subset $S'$ of $S$ can be contained in $S$, so the only simple set contained in $S$ is the trivial , empty set. And the empty set is defined to have measure $0$.

For the outer measure, you want to find the "smallest" set $T$ containing $S:=\mathbb Q \cap [0,1]$. As pointed above, by density of $\mathbb Q$ in $\mathbb R$ , no strict subset of $[0,1]$ can contain $S$. We then only have the option of having sets of the type $S'':=[0,1+e)$ covering $S$; we can rewrite $S'':=[0, 1+\frac{1}{n})$, and $m^*(S'')=1 + \frac{1}{n}$. The infimum of the measures over all $S''$ is then $ inf$ { 1+$\frac{1}{n}$ : n in $\mathbb N$ }, which is 1.

  • $\begingroup$ there is some backward bracket ")" missing, making the first paragraph a bit hard reading. The point "(defined as the sup of the measures..." missed the closing. $\endgroup$ – hhh Sep 9 '11 at 19:56
  • $\begingroup$ +1 the second paragraph is excellent, still trying decode the first one... $\endgroup$ – hhh Sep 9 '11 at 19:59
  • $\begingroup$ Thanks, hhh, I am having some technical difficulties with the 1st paragraph; I'm on it, tho. $\endgroup$ – gary Sep 9 '11 at 20:11
  • $\begingroup$ what about the situation when you have $\mathbb R \cap (0,1)$? $\mu ( \mathbb R \cap (0,1) ) = 0$? $\endgroup$ – hhh Sep 9 '11 at 20:44
  • $\begingroup$ $\mathbb R \cap (0,1)$ is $(0,1)$, which is a simple set. Do you mean $(\mathbb R-\mathbb Q) \cap (0,1)$? $\endgroup$ – gary Sep 9 '11 at 20:46
  • The article you linked to explains it clearly- the expression you posted is the measure of a single rectangle, and one may add up the measures of a countable number of these rectangles separately to find the total measure of their union if the rectangles are disjoint but otherwise, they may have overlaps and you have to either account for the overlaps somehow or settle for sub-additivity. These are standard defining properties of measures/outer measures, which I suggest you review.

  • Yes, the Jordan measure of a single point is 0. Just apply the definitions using a very small covering rectangles.

  • Yes the Jordan measure of $[0,1]$ is $1$. Take outer measures with the rectangles $[0,1+\epsilon) $ and inner measures with rectangles $[0,1-\epsilon)$.

  • $\bar{\mu} ( \{Q \cap (0,1) \} ) = 1$ because the rationals are dense in the reals, so covering the rationals in $[0,1]$ with semi-open rectangles nessitates covering $[0,1)$ as well. On the other hand, $\underline{\mu} ( \{Q \cap (0,1) \} ) = 0$ because the irrationals are dense in the reals, so any simple set you try to place inside $\{Q \cap (0,1) \} $ will intersect with irrational numbers which are not in the set, so the only simple set inside is the empty set. By definition, the measure of the empty set is $0$.

In response to your edits: We are just applying the definitions of outer and inner Jordan measure. Hopefully my 4th point above addresses your question. The inner and outer measure of (0,1) is indeed 1, you should try proving it from the definitions as an exercise.

  • $\begingroup$ I can create a cover $(\mp \epsilon, 1 \pm \epsilon)$ for $(0,1) \cup R$ so $\mu ( (0,1) ) = \lim_{\epsilon \rightarrow 0_{+}} (\mp \epsilon, 1 \pm \epsilon) = 1$. But only in $R$! The measure for $((0,1) \cup Q)$ is then a puzzle. I cannot yet understand why $\bar{\mu} = 1$ while $\underline{\mu} = 0$, thinking. $\endgroup$ – hhh Sep 9 '11 at 18:23
  • $\begingroup$ Do you mean that $\bar{\mu} ( \{[0,1] \cup Q \} ) = 1$ with the part "nessitates covering $[0,1)$ as well."? Now I am a bit puzzled actually with this one. I would call it undefined or why the other end is not needed to be $]$? How can you prove it? $\endgroup$ – hhh Sep 9 '11 at 18:35
  • $\begingroup$ No, I purposely meant the intersection, not the union. Try thinking about covering all the rationals in [0,1] by rectangles, no matter what rectangle you choose to cover each rational point, it will take up some non-zero length. And to do this for all rationals in that interval, ends up covering [0,1). $\endgroup$ – Ragib Zaman Sep 9 '11 at 18:39
  • 1
    $\begingroup$ No, I did not say those. Similar ideas to what I said in my 4th point can be applied to these other 3 cases if you think about them. In fact, you can address them all at once with the same logic: The inner measure of any subset of the rationals must be 0, because no simple sets expect the empty set can be contained in the rationals. I am sorry, but I must sleep now. I encourage any others reading this to please help hhh in my absence. $\endgroup$ – Ragib Zaman Sep 9 '11 at 18:58
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    $\begingroup$ hhh: $(0,1)$ is contained in $R$, so $(0,1)\cap R=(0,1)$ $\endgroup$ – gary Sep 9 '11 at 19:57

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