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I found this problem, and I can't get an answer to it:

Prove that there are subfields of $\Bbb{R}$ that are

a) non-measurable.

b) of measure zero and continuum cardinality.

I can't seem to imagine how to construct such subfields of $\Bbb{R}$.

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For (b), my idea would be to use the cantor set and have a look at the subfield of $\mathbb{R}$ generated by this set. –  Paŭlo Ebermann Jun 22 '11 at 22:21
    
@Beni Bogosel: Old result of Kuratowski, but may have used CH, I don't remember. –  André Nicolas Jun 22 '11 at 22:49
    
Are there some extension fields of $\Bbb{Q}$ which may turn out to be non-measurable? –  Beni Bogosel Jun 22 '11 at 22:56
    
The standard cantor set $E$ certainly doesn't work because $E - E$ contains an interval. You'll need to remove a lot more than middle thirds. –  Robert Israel Jun 22 '11 at 22:59
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@Robert, @George, so if this is a solution, then why not post a fully explained answer? –  JDH Jun 24 '11 at 20:15

3 Answers 3

up vote 8 down vote accepted

I expect direct constructions, but while we are waiting, let me at least offer a proof that both assertions are consistent with ZFC, since they can be obtained by forcing.

Statement (b) is true of the ground-model reals after adding a Cohen real. In the answer to this MO question, Martin Goldstern explains that the set of reals in the ground model $\mathbb{R}^V$ has measure $0$ in the forcing extension $V[c]$, but it is easy to see that it has size continuum there. So this is a subfield of $\mathbb{R}$ in the universe $V[c]$ which has measure $0$ and size continuum.

Statement (a) is true in the models mentioned in Andreas Blass' comment to my answer to this MO question, where it is explained that the set of constructible reals, which is certainly a subfield, can be non-measurable afte forcing over the constructible universe.

Perhaps one can turn both of these arguments into actual ZFC constructions by considering partially generic filters. Or perhaps there are easier direct constructions.

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Thank you very much for your quick answer. I will wait for some simpler direct constructions, if they are possible. –  Beni Bogosel Jun 22 '11 at 22:58
    
I will accept your answer, although it is not very simple to understand. Have you found any other simpler solutions? –  Beni Bogosel Jun 27 '11 at 21:33
    
Beni, thanks for accepting, although I still expect someone to post a direct construction (such as the Hausdorff dimension argument mentioned in the comments). So please feel free to unaccept later. –  JDH Jun 28 '11 at 11:30

Rather bizarrely, here's an answer to one of the questions, but I'm not sure which. But maybe it's a useful starting point.

Pick a transcendence basis of $B$ of $\mathbb{R}$ over $\mathbb{Q}$, any $b \in B$, and look at the field $F$ generated by $B\setminus \{b\}$. Since the translates of $F$ by elements of $\mathbb{Q}b$ are pairwise disjoint, $F$ is either null or nonmeasurable.

I suspect (hope?) that this is nonmeasurable, maybe after taking algebraic closure. If this is so, then a strategy for getting a null field could be tossing out more elements of $B$.

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You can try the answers (including mine) for a question at MathOverflow... http://mathoverflow.net/questions/27352/a-question-about-fields-of-real-numbers/27358#27358

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