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I have read that if we assume the continuum hypothesis then it can be proved or concluded tha there exist a set function μ that has the three following properties:

  • μ(A) is defined for each set A of real numbers
  • For an interval I, μ(I)=/I/
  • And if {En} is a sequence of disjoint sets then μ(UEn)=Σμ(Εn)

I cannot understand the connection between the existance of such a set function and the continuum hypothesis.

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    $\begingroup$ Assuming that everything I read on the internet is true, you have this backwards - CH implies that there does not exist such a $\mu$. This is clear if we believe Ulam's theorem: A measure defined on the power set of $\aleph_1$ which vanishes on singletons is equal to zero. math.stackexchange.com/questions/95964/… $\endgroup$ – David C. Ullrich Apr 24 '16 at 13:14
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    $\begingroup$ ... and this was exactly why Ulam proved the theorem: to show there was no (countably additive) extension of Lebesgue measure to all sets. $\endgroup$ – GEdgar Apr 24 '16 at 13:23
  • $\begingroup$ @David: This poses a unique problem, since you also read 5hie question on the internet, and therefore you assume that it is true... But you claim it isn't! :-P $\endgroup$ – Asaf Karagila Apr 24 '16 at 14:26
  • $\begingroup$ @AsafKaragila Heh. Actually I have no problem with an of the assertions in the OP. It may well be that he read what he says he read... $\endgroup$ – David C. Ullrich Apr 24 '16 at 14:45
  • $\begingroup$ For the record, I stated Ulam's theorem incorrectly. The theorem is that a finite measure defined on the power set of $\aleph_1$ which vanishes on singletons must be zero. This is false without the word "finite": For $E\subset\omega_1$ define $\mu(E)=0$ if $E$ is bounded above (in the standard order on $\omega_1$) and $\mu(E)=\infty$ if $E$ is unbounded. $\endgroup$ – David C. Ullrich Apr 24 '16 at 18:06
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Edit There was a major hole in the first version, fixed with the aid of a hint from hot_queen. See Below.

It goes the other way; CH implies that there is no such measure.

This follows from Ulam's theorem, which says that if $\mu$ is a finite measure defined on the power set of $\omega_1$ and $\mu$ vanishes on singletons then $\mu=0$.

Been wondering how to prove that; made this into an answer when I saw how simple it is. Say $\mu$ is a finite measure defined on the power set of $\omega_1$ which vanishes on the singletons. Consider the product measure $\mu\times\mu$ on $\omega_1\times\omega_1$. Let $A=\{(\alpha,\beta)\in\omega_1\times\omega_1:\alpha<\beta\}$. Tonelli shows that $$0=\mu\times\mu(A)=\mu(\omega_1)^2.$$

In fact the same proof works assuming just that $\mu$ is $\sigma$-finite. On the other hand, if we define $\mu(E)$ for $E\subset\omega_1$ by $\mu(E)=0$ if $E$ is bounded and $\mu(E)=\infty$ otherwise then $\mu$ is a non-zero measure defined on the power set of $\omega_1$ which vanishes on singletons; hence you can't ignore the hypotheses in Tonelli's theorem.


Below: hot_queen pointed out that I neglected to show that $A$ was measurable (wrt to the product $\sigma$-algebra). This is not that hard once you see how to do it; I needed a hint.

There exist sets $I_{n,j}\subset\Bbb R$ such that $$\{(x,y)\in\Bbb R^2:x=y\}=\bigcap_{n\in\Bbb N}\bigcup_{j\in\Bbb Z}(I_{n,j}\times I_{n,j});$$for example $I_{n,j}=[j/n,(j+1)/n)$. Since $\omega_1$ is equivalent to a subset of $\Bbb R$ it follows that there exist $I_{n,j}\subset\omega_1$ such that$$\{(\alpha,\beta)\in\omega_1^2:\alpha=\beta\}=\bigcap_{n\in\Bbb N}\bigcup_{j\in\Bbb Z}(I_{n,j}\times I_{n,j}).$$Hence if $J_{n,j}$ is the set of countable ordinals greater than or equal to the smallest element of $I_{n,j}$ we have $$\{(\alpha,\beta)\in\omega_1^2:\alpha\ge\beta\}=\bigcap_{n\in\Bbb N}\bigcup_{j\in\Bbb Z}(J_{n,j}\times I_{n,j}).$$

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  • $\begingroup$ You also need to show that your set $A$ is measurable in the product measure before you could apply Fubini. This is true but not entirely obvious. $\endgroup$ – hot_queen Apr 24 '16 at 19:49
  • $\begingroup$ @hot_queen Sigh. Good point... $\endgroup$ – David C. Ullrich Apr 24 '16 at 20:27
  • $\begingroup$ @hot_queen I'm missing the essential trick. Hint? $\endgroup$ – David C. Ullrich Apr 24 '16 at 23:21
  • $\begingroup$ Start by showing that the graph of any function from $\omega_1 \to \mathbb{R}$ is in the sigma algebra generated by rectangles of the form $A \times U$ where $A $ is arbitrary and $U \subseteq \mathbb{R}$ is open. $\endgroup$ – hot_queen Apr 24 '16 at 23:53
  • $\begingroup$ @hot_queen If $I_{n,j}=((j-1)/n,(j+1)/n)$ and $E_{n,j}=f^{-1}(I_{n,j})$ then the graph of $f$ is $\bigcap_{n\in\Bbb N}\bigcup_{j\in\Bbb Z}(E_{n,j}\times I_{n,j})$. Not that I see how that helps (but gimme a little time on that). $\endgroup$ – David C. Ullrich Apr 25 '16 at 0:21

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