# Measure of the same set in different models of ZF

Let $A$ be a definable subset of $\mathbb{R}$ in $\mathsf{ZF}$, and let $\mathcal{M},\mathcal{N}\models\mathsf{ZF}$ such that $A$ is lebesgue measurable in both models.

Is $\mu^\mathcal{M}(A^\mathcal{M})=\mu^\mathcal{N}(A^\mathcal{N})$?

Are there any conditions (applied to $A$ or $\mathcal{M},\mathcal{N}$) under which we know more about this question?

• What exactly do you mean "definable" here? – Asaf Karagila Aug 4 '14 at 16:30
• I meant in the sense of first-order formula with one free variable that defines $A$ as a subset of $\mathbb{R}$. I actually don't care about the definability, but I can't see how to ask about $A$ in different models otherwise. – Shay Ben Moshe Aug 4 '14 at 16:32
• Not in the way you stated it. The usual counterexample is to define $A$ to be empty or the whole line depending on whether $\mathsf {CH}$ holds. What you want instead is to have a "robust" description, in which case the answer is yes. The problem is to formalize this "robustness". For Borel sets and their continuous images, we can exhibit "codes" that describe how the set is made up starting with basic open sets. Beyond this, we do not have entirely satisfactory codes and the problem becomes set-theoretic. – Andrés E. Caicedo Aug 4 '14 at 16:33
• Here is one and here is another (which might be less duplicate, but also relevant). – Asaf Karagila Aug 4 '14 at 16:36
• The strongest results we have require that we assume large cardinals, and restrict our attention to models and their forcing extensions, the codes being what we call term relations. You can read about this in the Feng-Magidor-Woodin paper on universally Bare sets, and in Steel's first paper on the derived model theorem. This is now part of the core model induction machinery. – Andrés E. Caicedo Aug 4 '14 at 16:38

Even when $\mathcal{M, N}$ are transitive models of $ZFC$ with same ordinals and cardinals, a nice counterexample is given by the set $L$ of constructible numbers which has a $\Sigma^1_2$ definition. One way to see this is by adding a Cohen real to a model of $V = L$.
• Perhaps it should be added that the set goes from full measure in $L$ to null measure in $V$. It's what the expert probably don't call "Full to null" (because it doesn't quite rhyme in American English...) – Asaf Karagila Aug 4 '14 at 17:25