Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question

1 Answer 1

up vote 19 down vote accepted

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.

share|improve this answer
    
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 1 '11 at 13:28
    
I added a more concrete example. –  Joel David Hamkins Mar 1 '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". –  Matteo Mio Mar 1 '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 1 '11 at 20:42
3  
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 3 '11 at 1:55

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.