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While reading David Williams's "Probability with Martingales", the following statement caught my fancy:

Every subset of $\mathbb{R}$ which we meet in everyday use is an element of Borel $\sigma$-algebra $\mathcal{B}$; and indeed it is difficult (but possible!) to find a subset of $\mathbb{R}$ constructed explicitly (without the Axiom of Choice) which is not in $\mathcal{B}$.

I am curious to see an example of such a subset.

t.b.
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Sasha
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1 Answers1

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I would like to write up an answer to my question, although I am bringing no input of my own, but compiling from various links provided in comments.

  1. There is a model of ZF in which the set of all real numbers is a union of countably many countable sets. This is theorem 10.6 on page 142 from the book of Tomas J. Jech, "Axiom of Choice".
    This implies that "all sets of reals are Borel" within this model. [Added However, in such a model, it is impossible to define countably additive Lebesgue measure. Thanks to @AsafKaragila for pointing this out.]

  2. Without the use of AC, one can construct a complete analytic set (i.e. a subset of $\mathbb{R}$ that is a continuous image of the space of irrational numbers, $\mathbb{R}\backslash \mathbb{Q}$). One can not prove that this set is non-Borel (here the model of ZF is different from that in item 1), without invoking the axiom of choice.

  3. An example of such an analytic set is due to Lusin, and described in a PlanetMath article by George Lowther, who also contibutes to this site. The set $S$ is defined as a set of continued fractions: $$ S = \{x \in \mathbb{R} : x = [a_0, a_1, a_2, \ldots ], \exists 0<i_1 < i_2 < \cdots, \forall_{k \geqslant 1} a_{i_k} | a_{i_{k+1}} \} $$ The set is Lebesgue measurable (the measure zero, right?), but not Borel measurable.

Tristan Phillips
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Sasha
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  • @AndresCaicedo I would appreciate your comment on this answer of mine, as it draws, in part, from your post on MO. Thanks. – Sasha May 03 '12 at 20:54
  • @GeorgeLowther Could you please look over this post, making sure that I am understanding your PlanetMath article on Lusin's example of Lebesgue measurable, but not Borel measurable set. Thank you. – Sasha May 03 '12 at 20:57
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    Sasha, indeed it is possible to have that the real numbers are a countable union of countable sets, and in such model indeed every set is Borel, in fact the Borel rank is finite. In such model, however, one cannot fully develop the notion of a countably additive Lebesgue measure. – Asaf Karagila May 03 '12 at 21:10
  • @AsafKaragila Thank you. This helps with understanding. I have expanded the answer to add your observation. By the way, if you happen to know, how would I make my answer a community wiki? Should I do this at all? – Sasha May 03 '12 at 21:20
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    If you wish to make the answer CW simply click the checkbox when editing it (should be between the edit box and the preview). Whether or not you should, I don't know. It is up to you... :-) – Asaf Karagila May 03 '12 at 21:22
  • I do not understand your comment in item 2. There are no models here, you have an explicit construction of a complete analytic set. Do you simply mean that item 2 holds in general, regardless of whether you assume the pathological situation of item 1, or otherwise? – Andrés E. Caicedo May 03 '12 at 21:27
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    A slightly more subtle thing is that one can develop a version of Lebesgue measure, even in the pathological situation described in item 1. The catch is that, rather than arguing about the sets themselves, we argue about their "codes". (There is a natural way to assign codes to Borel sets, that describe how they are constructed from basic open sets.) I am not claiming it is sensible to develop Lebesgue theory this way, of course. I haven't really seen this written in detail anywhere. – Andrés E. Caicedo May 03 '12 at 21:30
  • @AndresCaicedo Yes, I meant that in item 2 we no longer work with the model of ZF that was described in item 1. Thanks for your comments. – Sasha May 03 '12 at 21:38
  • @Andres: Have you looked in Fremlin's book? He has quite a chapter about measure without choice and quite a lot on Borel code-based measures. – Asaf Karagila May 03 '12 at 21:46
  • @AsafKaragila It may be there, you are right. I'll have to check. – Andrés E. Caicedo May 03 '12 at 21:50
  • @AsafKaragila The page on PlanetMath uses $a_i = \prod_{k=1}^i d_k$, this way $a_i$ divides $a_j$ for $j>i>0$. – Sasha May 03 '12 at 21:54
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    Actually we are both wrong. The set is made of the fractions $[d_0;d_1,\ldots]$ such that there is an infinite subsequence $i_k$ for which $d_{i_k}$ is a divisor of $d_{i_{k+1}}$ for all $i$. This is a notably different condition than both our statements. (I removed my previous comment) – Asaf Karagila May 03 '12 at 21:57
  • @AsafKaragila Yes, you are correct, of course. Thanks! The set I was talking about is a lot thinner, since it is $a_i | a_j$ for any $j>i>0$. – Sasha May 03 '12 at 22:04
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    @AndresCaicedo: Fremlin in volume 5II of his measure theory has chapter 56 devoted to measure theory without or with little choice and enters quite a bit of detail for Borel-codable sets; that chapter also has a section on what happens under determinacy, so this might be worth a look given your interests. There are ample references there, too. – t.b. May 04 '12 at 19:09