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Hi guys,

is it possible to change the probability of an event via forcing? More precisely, is there an innocent looking question on the probability of "something" whose answer is independent of ZFC?

All the best, Sebastian

sebastian
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1 Answers1

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There are several issues.

On the one hand, any set can be made countable by forcing, and this process will certainly affect the measure of the set, if it did not have measure zero in the ground model.

But in the context of the Lebesgue measure on the reals, say, it is natural to consider not the set itself, but the Borel description of the set, interpreted first in the ground model and then reinterpreted in the forcing extension. (For exampe, the "unit interval" of $V$ is not necessarily the same as the unit interval of a forcing extension $V[G]$, but we have a borel code that correctly picks out the unit interval when interpreted in any model of ZFC.) In this case, one gets a positive solution for preservation of measure. The reason is that the assertion that the measure of the set with Borel code $b$ is $x$ has complexity at most $\Sigma^1_2(b,x)$ and hence is absolute to all forcing extensions by the Shoenfield absoluteness theorem. In this sense, the measure of a measurable set cannot be affected by forcing.

Meanwhile, the use of other non-absolute descriptions can lead again to a negative answer, where the measure can be affected by forcing. For example, consider the set $X$ of all binary sequences $x$ whose sequence of digits is realized somewhere in the GCH pattern of cardinals, in the sense that there is an ordinal $\beta$ such that $x(n)=1$ iff $2^{\aleph_{\beta+n}}=\aleph_{\beta+n+1}$. If the Generalized Continuum Hypothesis holds, then $X$ has measure zero, since only one pattern is realized. But one can force the GCH pattern to realize all patterns, and so there are forcing extensions in which $X$ has full measure.

Here is another comparatively concrete example. Consider the set of reals that are constructible, in the sense of Gödel's constructible universe. This set has complexity $\Sigma^1_2$ in the descriptive set-theoretic hierarchy, which is just a step up from Borel. The set has full measure in the constructible universe, of course, but it is easily made to have measure zero in a forcing extension. Thus, the probability that a randomly chosen real number is constructible has an answer that is independent of ZFC, because in some models of set theory this probability is 1 and in others it is 0.

Joel David Hamkins
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  • thanks a lot. that is already in the direction I am looking for. Do you know of any non-absolute description that one could use that is still "simple"? More precisely, the example above is a non-absolute description by basically providing a direct link from the sequences to the cardinal patterns. is there something more "basic" or "elementary"? I understand that it has to be somewhat more complex than the Borel code.

    thx a lot again!

     – sebastian Mar 01 '11 at 13:28
  • I added a more concrete example. – Joel David Hamkins Mar 01 '11 at 14:06
  • @Joel: I have the following naive question: is it true that the the set you defined is at least measurable in all models of set theory? Or in other words, is it possible to re-state your conclusion as "because in some models of set theory this probability is 1 and in others it is 0, and in others is not defined at all". –  Mar 01 '11 at 15:22
  • Matteo, that is an interesting question. More generally, if $V\subset V[G]$ is any forcing extension, then must $\mathbb{R}^V$ be measurable in $V[G]$? Must it always have either full measure or measure $0$? Hmmmmm... – Joel David Hamkins Mar 01 '11 at 20:42
  • The set of constructible reals is all the reals and hence Borel in $L$ itself, but in forcing extensions $L[G]$ it has complexity $\Sigma^1_2$, and I believe this bound can be realized (one must not collapse $\omega_1^L$), so it can happen that a Borel set is made non-Borel by forcing. But this doesn't fully answer my question.... – Joel David Hamkins Mar 02 '11 at 13:55
  • Right. Of course it is consistent that every $\Sigma^{1}{2}$ set is measurable. I'm quite interested in this thing, because in my research i'm working with a $\Delta{2}^{1}$ set. To work with it I need to assume something (such as Martin's Axiom + \neg CH, say) to force its measurability (MA + \negCH, implies that every $\Sigma^{1}_{2}$ set is measurable).

    So I'm interested to learn if there are ways to prove that it is measurable in ZFC alone, perhaps showing that it is a "constructible set" (I'm not an expert at all in this area, so i'm just guessing).

     –  Mar 02 '11 at 16:25
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    There are models of set theory in which the set of constructible reals is not Lebesgue measurable. If I remember correctly, examples include the models obtained from $L$ by adjoining a random real, a Laver real, a Sacks real, or a Miller real. These models satisfy CH, but countable support iterations of Laver, Miller, or Sacks forcing give models where the continuum has cardinality $\aleph_2$ yet the constructible reals are again non-measurable. And you can get the same result with even larger continuum by forcing with a big measure algebra to add a lot of random reals. – Andreas Blass Mar 03 '11 at 01:55
  • Thanks, Andreas, do you know where we can find the details? – Joel David Hamkins Mar 03 '11 at 02:24
  • Gunter Fuchs and I observed that if the ground model reals are ever measurable in a forcing extension $V[G]$ with a new real, then the measure of $\mathbb{R}^V$ must be $0$, since you can perform a Vitali-like argument by considering translates of the ground model unit interval by a new real and all its powers, which must be disjoint. – Joel David Hamkins Mar 04 '11 at 23:23
  • Joel, I'd expect that the details concerning my comment are in the Bartoszynski-Judah book "Set Theory: On the Structure of the Real Line". The case of Laver forcing is in a paper of Pawlikowski, "Laver's forcing and outer measure" in the proceedings of the 1992-94 BEST. He says the result is due to Woodin, rediscovered by Judah and Shelah, but "[Woodin's proof] is unpublished and the proof in [the Judah-Shelah paper on the Kunen-Miller chart] is somewhat difficult to follow." – Andreas Blass Mar 09 '11 at 16:34