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.

  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ ... 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

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 $$\{(\alpha,\beta)\in\omega_1^2:\alpha\ge\beta\}=\bigcap_{n\in\Bbb N}\bigcup_{j\in\Bbb Z}(J_{n,j}\times I_{n,j}).$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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