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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 …

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 …

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 …

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 …

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 …

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^* …

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 …

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 …

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 …

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 …

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