Let $\mathcal{A}$ denote the collection of all subsets A of an uncountable set $\Omega$ for which either A or $A^c$ are countable. Let $\mu(A)$ denote the cardinality of A. Define $\phi(A)$ equal to $\emptyset$ or $\infty$ according as $A$ is countable or uncountable. Show that $\phi ≪ \mu$. Then show that the Radon-Nikodym theorem fails.

This is an assignment. I found that this problem is not rigorous unless in this case we treat $+\infty$ as a number; and this equation, $+\infty+\infty=+\infty$ holds where we do not distinguish the cardinality of all the rational number, $\aleph_0$, and the cardinality of real number, $\aleph_1$, and so on.

If we treat $\infty$ as a number, it is easy to show that $\mathcal{A}$ is a $\sigma$-field, $\mu$ and $\phi$ a measure and $\phi ≪ \mu.$

My question is: Does this make sense $\aleph_0+\aleph_1+\aleph_2$? (Set theory is not my suit.)

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    $\begingroup$ Sure, you can talk about $\aleph_0 + \aleph_1 + \aleph_2$. It's equal to $\aleph_2$. But I don't think that's the correct way to approach this problem. $\endgroup$ – Zhen Lin Mar 13 '11 at 16:53
  • $\begingroup$ In this case, could we treat $\aleph_i$ as the same number,$\infty$? $\endgroup$ – Jun Deng Mar 13 '11 at 17:14
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    $\begingroup$ You seem a bit confused. The $\infty$ of the extended real numbers -- i.e., such that in real analysis the notion of "real-valued function" is relaxed to functions which may take values in $[-\infty,\infty]$ -- is not the same notion as the cardinality of a(ny) infinite set. In particular this problem doesn't have anything to do with cardinal arithmetic. $\endgroup$ – Pete L. Clark Mar 13 '11 at 17:32
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    $\begingroup$ $\aleph_1$ is not necessarily the cardinality of the real numbers! $\endgroup$ – Henno Brandsma Mar 13 '11 at 17:36
  • $\begingroup$ Thanks Henno. Sure. Tricky Cardinality of the continuum. $\endgroup$ – Jun Deng Mar 13 '11 at 18:34

A measure (in your case) takes values in $[-\infty, +\infty]$ or even $[0,\infty]$ probably. The extended reals do have the arithmetric "rule" or convention that $+\infty + x = \infty$ for $x$ a normal real or $\infty$, while $+\infty + (-\infty)$ is undefined. This means that the description of the measure $\mu$, the so-called counting measure, must mean that any infinite set $A$, either countable or uncountable, gets $\mu(A) = +\infty$; this is what is meant by the fact that the measure is the cardinality: we only distinguish the finite ones, and all infinite ones get value $\infty$. The addition convention ensures that this will work as a measure, as you said. The values thus are not cardinals, but extended reals.

With this information I hope you can figure out the question.

  • $\begingroup$ Thanks, Henno. This does help me! $\endgroup$ – Jun Deng Mar 13 '11 at 18:36

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