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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.

Related

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

2 Answers 2

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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

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.

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    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.2011-09-09
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    +1 the second paragraph is excellent, still trying decode the first one...2011-09-09
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    Thanks, hhh, I am having some technical difficulties with the 1st paragraph; I'm on it, tho.2011-09-09
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    what about the situation when you have $\mathbb R \cap (0,1)$? $\mu ( \mathbb R \cap (0,1) ) = 0$?2011-09-09
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    $\mathbb R \cap (0,1)$ is $(0,1)$, which is a simple set. Do you mean $(\mathbb R-\mathbb Q) \cap (0,1)$?2011-09-09
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    The Jordan measure of boundary is zero IFF Jordan measurable (my book, not in English). Closed sets are Jordan measurable IFF their characteristic function $\chi_{A}$ is Rieman integrable. My book states that Jordan measure is a "closed set measure", apparently meaning that Jordan measure is zero for all open sets or semi-open-sets -- they are still Jordan-measurable but they are zero, by definition. So $\mu (\mathbb R - \mathbb Q) \cap (0,1) ) = 0$ and $\mu ( (0,1) ) = 0$, by definition. I sense here that my book misses points here and there.2011-09-09
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    But $\mu ((0,1) - ( \mathbb R - \mathbb Q) ) = 1$ because endpoints are rational, the boundary is zero and the inner measure is 1 (not hitting irrational). Does it make any sense?2011-09-09
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    I'm sorry, I will be busy for a while; would you mind waiting till tomorrow?2011-09-09
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    sure, no hurry. I will be doing exercises.2011-09-09
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    You can use, to show $m((0,1)-(\mathbb R- \mathbb Q) \cap(0,1) )=1$ by using the fact that $A:=(0,1) /cap \mathbb Q$ and $B:=(0,1)\cap (\mathbb R-\mathbb Q)\cap (0,1)$ are disjoint, and their union is (0,1), so that , by additivity, $m(A\cup B)=m(A)+m(B)=m(0,1)=1$.2011-09-10
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    $\underline{\mu} ( \{Q \cap (0,1) \} ) = 0$ where $(0,1) = (0,\frac{1}{3})\cup [\frac{1}{3}, \frac{2}{3}] \cup (\frac{2}{3}, 1)$ where the sub-sets are disjoint. We know that $\mu(A\cup B) = \mu(A) + \mu(B)$ for disjoint sets. So $$\mu \left((0,\frac{1}{3})\cup [\frac{1}{3}, \frac{2}{3}] \cup (\frac{2}{3}, 1)\right) = \mu (0,\frac{1}{3})) + \mu( [\frac{1}{3}, \frac{2}{3}] ) + \mu (\frac{2}{3}, 1) = 0 + \frac{1}{3} + 0$$ but it was zero, contradiction. What is wrong here?2011-09-10
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    The question is now [here](http://math.stackexchange.com/questions/63352/jordan-measure-with-semi-closed-sets).2011-09-10
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  • 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.

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    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.2011-09-09
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    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?2011-09-09
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    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).2011-09-09
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    Now I understood the last bullet point and reading teh wikipedia again, the Jordan measure is just a generalization of other spatial things such as lenght, area and volume. But I cannot yet understand the semiclosed interval to be 1, thinking. The semi-closed interval does not cover the irrational in the other end, it cannot be 1. Please, elaborate on this point.2011-09-09
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    sorry, typo. I meant intersection of course, $\bar{\mu} ( \{[0,1] \cap Q \} ) = 1$. I am puzzled by the part `"nessitates covering $[0,1)$ as well."`? Semi-closed interval?2011-09-09
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    No, the number 1 is rational! It must be in the cover or otherwise you are not covering all points. We know that $\bar{\mu} ([0,1] \cap Q) = 1$, $\bar{\mu} ([0,1) \cap Q) = 1$ and $\bar{\mu} ((0,1] \cap Q) = 1$. Are you really claiming with the penultimate premises $$\underline{\mu} ( [0,1)\cap Q) = \underline{\mu} ( (0,1]\cap Q ) = 1$$ so $$\mu ( [0,1)\cap Q) = 1$$ with semi-closed sets while $$\underline{\mu} ([0,1] \cap Q) = 0$$ so $$\mu ([0,1] \cap Q) = 0$$?2011-09-09
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    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.2011-09-09
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    Any subset of $Q$ such as $\{ Q \cap [0,1) \}$ must have zero-measure because the lower measure is zero -- because the cover does not have the irrational end points so not covering. $\mu ( [0,1) \cap Q) = 0$ because the lower measure is zero with empty cover. I can understand it this way, I think I am probably just misunderstanding the writing.2011-09-09
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    Do I infer right that $\mu( (0,1) \cap R ) = 1$, where $R$ is real numbers? Because the lower cover will now contain all points even the imaginary points, is that correct?2011-09-09
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    The zero measure of this set is because the largest (unique) simple subset contained in $\mathbb Q$ is the empty set; consider any simple set $S:=[a,a+e)$ containing some rational $q$, will not be contained in $\mathbb Q$, since both rationals and irrationals are dense in $\mathbb R$. For the upper measure, the smallest simple set containing $\mathbb Q \cap [0,1]$ is $[0,1+e)$. For more rigor, we have that $[0,1+\frac{1}{n})$ covers $\mathbb Q \cap [0,1]$, for all n in $\mathbb N$ . The measure of each of these sets is $1+\frac{1}{n}$, so the outer measure is $inf{1+\frac {1}{n}}$ for all n.2011-09-09
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    Let's consider $A = \cup_{k=1}^{\infty} [3*2^{-k}, 8*2^{-k}]$. So $lim_{p \rightarrow \infty} \cup_{k=1}^{p} [3*2^{-k}, 8*2^{-k}] = [0,4]$ or is it just $(0,4]$? So is the measure in this example $4$ or $0$?2011-09-09
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    @gary: sorry but do you mean $\mu ((0,1) \cap R) =0$?2011-09-09
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    hhh: look at my answer below, and let me know if it helps.2011-09-09
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    hhh: $(0,1)$ is contained in $R$, so $(0,1)\cap R=(0,1)$2011-09-09
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    @gary: is $\mu ( (0,1) \cap R) = \mu ( (0,1) \cap R) = \mu ( (0,1) ) = 0$ so is the Jordan measure zero for all open sets? Is the Jordan measure zero for semi-closed sets?2011-09-09