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On the page 43 of Real Analysis by H.L. Royden (1st Edition) we read: "(Ideally) we should like $m$ (the measure function) to have the following properties:

  1. $m(E)$ is defined for each subset $E$ of real numbers.
  2. For an interval $I$, $m(I) = l(I)$ (the length of $I$).
  3. If $\{E_n\}$ is a sequence of disjoint sets (for which $m$ is defined), $m(\bigcup E_n)= \sum m (E_n)$."

Then at the end of page 44 we read : "If we assume the Continuum Hypothesis (that every non countable set of real numbers can be put in one to one correspondence with the set of all real numbers) then such a measure is impossible," and no more explanation was given.

Now assuming the Continuum Hypothesis I am not able to see why such a measure is not possible. Would you be kind enough to help me?

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    $\begingroup$ Please do not write all-caps. You should edit this to fix it... $\endgroup$ – Mariano Suárez-Álvarez Jan 3 '12 at 2:20
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    $\begingroup$ @Kamram: To explain, on the web, writing in all capitals is considered shouting. You are basically yelling at people. $\endgroup$ – Arturo Magidin Jan 3 '12 at 2:21
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This is a consequence of Ulam's theorem: If a finite countable additive measure $\mu$ is defined on all subsets of a set of cardinality $\aleph_1$ (the first uncountable cardinal) and vanishes on all singletons, then it is identically zero.

This is the version of the theorem from the book of Bogachev (1.12.40), the original result is a bit stronger. Anyways, under CH, $\mathbb{R}$ has cardinality $\aleph_1$, so the theorem applies to $\mathbb{R}$. I'm pretty sure this is what Royden refers to.

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    $\begingroup$ I would argue that the form of Royden's remark is rather Banach-Kuratowski's negative solution to Lebesgue's measure problem under CH, see their article Sur une généralisation du problème de la mesure on which Ulam's paper expands. Note also that relaxing the countable additivity condition 3. to finite additivity, there is such a function $m$, as was shown earlier by Banach: Sur le problème de la mesure. $\endgroup$ – t.b. Jan 3 '12 at 5:59
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    $\begingroup$ The result can also be found in Oxtoby book. $\endgroup$ – sdcvvc Jan 3 '12 at 7:10
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I refer you to the MathOverflow question "Does pointwise convergence imply uniform convergence on a large subset?" for proofs and references for the following:

Assuming the Continuum Hypothesis, there exists a sequence of real-valued functions on $\mathbb R$ that converges pointwise to a function on $\mathbb R$ but does not converge uniformly on any uncountable set.

Suppose there were a function $m:\mathcal{P}(\mathbb R)\to[0,\infty]$ satisfying the three named conditions. Then $m$ is a measure on $\mathbb R$ for which every subset of $\mathbb R$ is measurable. Let $(f_n)$ be a sequence of real-valued functions on $[0,1]$ that converges pointwise to some function. By Egorov's theorem, there is a set $E\subseteq [0,1]$ with $m(E)>0$ such that $(f_n)$ converges uniformly on $E$. The fact that $m(E)>0$ implies that $E$ is uncountable. Since $(f_n)$ was arbitrary, this implies by the above cited result that the Continuum Hypothesis does not hold.


In light of Michael's answer, I want to mention that the hypothesis 2, that $m(I)=l(I)$ for each interval $I$, could be replaced by the following two properties:

(a) For all $x\in\mathbb R$, $m(\{x\})=0$.

(b) There exists $X\subseteq\mathbb R$ such that $0<m(X)<\infty$.

Then the same argument as above would apply with such an $X$ in place of $[0,1]$.

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