Gently changing measure

Question

This question was asked and bountied on MSE without answer, so I'm porting it here:

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There's an easy way to change the measure of a set of reals by moving to a larger universe: simply make $\mathbb{R}$ of the ground model be null. Conversely, it's not hard to show that if $M\subseteq N$ are transitive models of ZFC and $A\in\mathcal{P}(\mathbb{R})^M$ is measurable in $M$ and has positive measure in $N$, then $m(A)^M=m(A)^N$, so at least for "nice" sets that's the only way measure can change via this process. The answer there is stated for forcing extensions only, but that's unnecessary.

For non-measurable sets things are more complicated. However, the construction given there still somewhat fits the "nullify-or-leave-unchanged" pattern: the anomolous set is built from two pieces such that we can make one null while not changing the outer measure of the other. I'm interested in whether this is optimal:

Question 1: Can there be a pair $M\subseteq N$ of transitive models of ZFC and a set of reals $A\in M$ such that $(i)$ $\mu^*(A)^M>\mu^*(A)^N$ but $(ii)$ there is no partition $A=B\sqcup C$ with $B,C\in M$ such that $\mu^*(B)^N=0$ and $\mu^*(C)^N=\mu^*(C)^M$?

I suspect the answer is no. Annoyingly, I haven't been able to make any progress on this, and in particular I can't even rule out the following extreme (and "obviously" ridiculous) possibility:

Question 2: Can there be a pair $M\subseteq N$ of transitive models of ZFC such that no non-null $A$ in $M$ is null in $N$ but some $A$ in $M$ has $\mu^*(A)^M\not=\mu^*(A)^N$?

Answer 1

To question 1, there is such a pair. This is a minor reworking of Ashutosh's example you linked.

Start with $L,$ add an $\omega_1$-sequence of random reals $X=\langle r_{\alpha}: \alpha<\omega_1 \rangle,$ then a Cohen real $c,$ then another $\omega_1$-sequence of random reals $Y=\langle s_{\alpha}: \alpha<\omega_1 \rangle,$ then another random real $t.$ Let $N=L[X][c][Y][t]$ be the final model. Identify the reals in $X$ and $Y$ as elements of $[0,1],$ and $t$ as a subset of $\omega.$ In $N,$ $\mu^*(X)=0$ and $\mu^*(Y)=1.$

We now work in $L[X][Y][t],$ noting that $X,$ $Y,$ and $t$ are mutually random over $L.$ Let $\langle b_n \rangle$ be the increasing enumeration of $t,$ and $\langle c_n \rangle$ the increasing enumeration of $\omega \setminus t.$

We use $t$ to interweave $X$ and $Y.$ Precisely, define $\langle u_{\alpha}: \alpha<\omega_1 \rangle$ by letting $r_{\omega \alpha + n} = u_{\omega \alpha + b_n}$ and $s_{\omega \alpha + n} = u_{\omega \alpha + c_n}.$ Applying permutation invariance of random forcing in $L[t],$ we have that $u$ is a random $\omega_1$-sequence of random reals over $L[t]$ and $t \not \in L[u].$

Let $M=L[u] \subset L[X][Y][t] \subset N,$ and let $A=\{2^{-n-1}(1+u_{\omega \alpha +n}): n<\omega, \alpha<\omega_1\}.$ Then $\mu^*(A)^M=1$ and $\mu^*(A)^N = \sum_{n \in \omega \setminus t} 2^{-n-1} \in (0,1).$

Suppose there is a partition $A = B \sqcup C$ in $M$ as in the description of question 1. Then $t = \{n<\omega: \mu^*( C \cap [2^{-n-1}, 2^{-n}])^M = 0\} \in M,$ contradiction.

I think "morally" this counterexample doesn't really violate your intuition, in that we still have in $N$ a partition of $A$ into one set of "preserved points" and another set of "nullified points." A natural follow-up question (which I don't know the answer to) is whether there can be $A \in M \subset N$ such that (i) $\mu^*(A)^M > \mu^*(A)^N$ and (ii) in $N,$ every point in $A$ is a Lebesgue density point of a minimal measurable cover of $A.$

Answer 2

Here's a sketch of an attempt at negative answer for #2. First, we may assume that our measure space is the Cantor space $2^{\omega}$, and that $\mu^{*}(A)^{M} = 1$. Working in $M$, $(2^{\omega})^{\omega}$ is isomorphic to $2^{\omega}$ and $\mu^{*}_{\omega}(A^{\omega}) = 1$, letting $\mu^{*}_{\omega}$ be the corresponding version of outer measure.

I would think that this last claim is standard, but I haven't been able to find it anywhere. One argument for it would be to take the construction of the $\omega$-size product in Theorem 254F of Fremlin (https://www1.essex.ac.uk/maths/people/fremlin/chap25.pdf) and apply it to measure space on $A$ induced by its being a subspace of $2^{\omega}$ of full outer measure. Another argument would be to take a supposed closed set $C$ of positive measure in $(2^{\omega})^{\omega}$ disjoint from $A^{\omega}$ and, using Fubini's Theorem, iteratively choose the members of a sequence $\langle a_{n} : n \in \omega \rangle \in A^{\omega} \cap C$.

If this is right then we seem to be done: if $\mu^{*}(A)^{N} < 1$, then $\mu^{*}_{\omega}(A^{\omega})^{N} = 0$.