I want to compute the lebesgue outer measure of $(0,1) \cap \mathbb{Q}^c$ where $\mathbb{Q}$ is the set of rational numbers.

$(0,1) = \{(0,1) \cap \mathbb{Q}\} \cup \{(0,1) \cap \mathbb{Q}^c\}$

By sub-additivity of lebesgue outer measure, we have $m^*\{(0,1)\} \leq m^*\{(0,1) \cap \mathbb{Q}\} + m^*\{(0,1) \cap \mathbb{Q}^c\}$

$\Rightarrow m^*\{(0,1) \cap \mathbb{Q}^c\} \geq m^*\{(0,1)\}(=1)- m^*\{(0,1) \cap \mathbb{Q}\}(=0)$

$\Rightarrow m^*\{(0,1) \cap \mathbb{Q}^c\} \geq 1$

Again by monotonicity of lebesgue outer measure, $m^*\{(0,1) \cap \mathbb{Q}^c\} \leq m^*\{(0,1)\}(=1)$

$\Rightarrow m^*\{(0,1) \cap \mathbb{Q}^c\} \leq 1$

Thus we have $m^*\{(0,1) \cap \mathbb{Q}^c\}=1$, as expected.

$\textbf{My problem:}$ I encountered a theorem that states: $m^*(U)=m_{*,(J)}(U)$ for all open sets $U$. Now $(0,1) \cap \mathbb{Q}^c$ seems to be an open set. Since $\mathbb{Q}$ is countable, so is $(0,1) \cap \mathbb{Q}$. Then we can write $(0,1) \cap \mathbb{Q}$ as $\{q_1,q_2, \ldots \}$. Thus we can write $(0,1) \cap \mathbb{Q}^c$ as $(0,q_1) \cup (q_1,q_2) \cup \cdots $, i.e. as countable union of open sets. Hence $(0,1) \cap \mathbb{Q}^c$ is open. Now $m_{*,(J)}\{(0,1) \cap \mathbb{Q}^c\}=0$. Then $m^*\{(0,1) \cap \mathbb{Q}^c\}=0$ (!!)

I maybe (must be!) completely off the track and I apologize in advance. Any clarification would be much appreciated!


$(0,1)\cap\mathbb Q^c$ is not open. Your attempt to write it as a union of open intervals (where your use of $\cap$ instead of $\cup$ is presumably just a typo) incorrectly presumes that $q_1<q_2<\dots$.

  • $\begingroup$ Isn't the issue not the assumption that they're ordered but the fact that rationals are dense in the reals? Any one of the OPs intervals is going to contain an extra rational number, and thus is not contained in $\mathbb{Q}^c$ $\endgroup$ – leibnewtz Dec 10 '16 at 18:13
  • $\begingroup$ @leibnewtz I agree that what you wrote describes an error in the OP's argument, but before that error, there was also an error in writing intervals like $(q_1,q_2)$ when $q_1$ might be bigger than $q_2$. So I"d summarize the situation as: The intervals written in the question won't all exist so their union doesn't make sense and the ones that do exist contain rational numbers. $\endgroup$ – Andreas Blass Dec 10 '16 at 18:24
  • $\begingroup$ What prevents us from writing them in ascending linear order? After some thought I realized that perhaps density also prevents any enumeration of the rationals to be ordered compatibly with the order on $\mathbb{R}$ $\endgroup$ – leibnewtz Dec 10 '16 at 18:27
  • $\begingroup$ Thanks! I think writing them in order is okay, but yeah, denseness of \mathbb{Q} in \mathbb{R} kicks the idea of having such intervals. My bad. $\endgroup$ – Akira Sengupta Dec 10 '16 at 18:49

Since $\mathbb{Q}$ is dense in the real numbers, any open interval $(a,b)$ contains a rational. It is therefore impossible to obtain an open set consisting entirely of irrational numbers.

  • $\begingroup$ Thanks a lot :) the thought of having such open intervals were crazy. $\endgroup$ – Akira Sengupta Dec 10 '16 at 18:47

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