# Why has Sacks' “Measure-theoretic uniformity” not been more influential?

In the 1969 paper "Measure-theoretic uniformity in recursion theory and set theory," Trans. Amer. Math. Soc. 142 1969 381–420, Sacks gave a measure-theoretic approach to several results previously obtained by forcing. For example, he showed that, when countably many random sets of natural numbers are added to a countable model of ZFC, the result with probability 1 is a model of ZF+DC+all sets of reals are measurable.

This seems to me a very enlightening view of Solovay's theorem, yet Sacks' paper is almost unknown. (It has 10 citations on MathSciNet, compared with 98 for Solovay's.) Is there any good reason for the relative neglect of Sacks' paper?

• Is it really correct to contrast adding random reals with forcing? After all, random-real forcing simply is a kind of forcing, namely, forcing with the measure algebra, as opposed to the Cohen algebra. When one says with random-real forcing that a result holds "with probability one", one is really saying that every forcing extension by that forcing notion has that property. I haven't looked at that paper recently, but I am confused by your description of it, since I had understood that results of Shelah show that one needs an inaccessible cardinal to achieve the consistency of the theory. – Joel David Hamkins Feb 8 '13 at 23:41
• If you mean Theorem 4.26 in Sacks' paper, note that the existence of an extension is not the same as "all sets are Lebesgue measurable". – Asaf Karagila Feb 9 '13 at 1:26
• To add on the above, Solovay proved in 1965 the consistency of an extension of the Lebesgue measure without choice, which it what Sacks refers to (reference no. 29; 30) whereas Solovay's celebrated consistency proof which requires inaccessible was published in 1970, and is a whole other body of work. – Asaf Karagila Feb 9 '13 at 1:43
• Indeed it is not. – Andrés E. Caicedo Feb 9 '13 at 3:00
• What Sacks proves is different. He obtains models depending on a real parameter such that (with probability 1) the model satisfies ZF+DC+there is a translation invariant extension of Lebesgue measure to all sets of reals. This is not the same as saying that all sets are measurable. The measure he obtains is not regular, of course. – Andrés E. Caicedo Feb 9 '13 at 3:06