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ZFC refutes this principle. Let $\kappa=\text{non}(\mathcal{L}),$ i.e. the least cardinality of a set of reals of positive outer measure. Let $X \subset [0,1]$ be such that $|X|=\kappa$ and $\lambda^* …

The answer to your first question is yes, and the answer to your second question is no, under any of the multiple definitions of "measurable" in choiceless contexts. We will prove a theorem relating v …

Let $\mu$ be a total extension of Lebesgue measure. For $\lambda>0,$ we define another total extension of Lebesgue measure by $\mu_{\lambda}(X)=\frac{1}{\lambda}(\lambda X).$ We will show there is $\l …

This question is explored in great generality by Laczkovich in Laczkovich, Miklós, "Two constructions of Sierpiński and some cardinal invariants of ideals", Real Anal. Exch. 24(1998-99), No. 2, 663-67 …

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 \rang …

The answer is no. This is the main result of "Extensions of invariant measures on Euclidean spaces" by Ciesielski and Pelc.

In any model of ZFC where there is no such set for $c=1$ (e.g., Friedman's model mentioned in Gro-Tsen's answer), there is no $S \subset [0,1]^2$ such that all vertical slices $S_x$ are null and all h …

The answer to your second question is yes. Let $G=(\mathbb{Q}[\sqrt{2}], +).$ This is a countable abelian group, so $\mathbb{R}/G$ is a hyperfinite Borel equivalence relation. In particular, it embeds …

This is more of a long comment than an answer. The "right" notion of an unbounded set being measurable in ZF is less than clear. Suppose $\mathbb{R}$ is a countable union of countable sets. Let $\lang …

There is no such measure. Suppose toward contradiction $\mu$ is such a measure. Partition $\Omega$ by the mod finite equivalence relation $\sim_{\text{fin}},$ let $X \subset \Omega$ be a choice of rep …

The answer to your first question is no. We will show there is a null Vitali set in Cohen's first model $M.$ Recall that in $M,$ there is an infinite (but Dedekind finite) set $A$ of mutually Cohen re …

No. Suppose $\mathcal{F}$ is such a filter. Clearly each $X \in \mathcal{F}$ has positive measure intersection with every positive-length interval. Then neither $[0,1/2]$ nor $[1/2, 1]$ are in the fil …