Almost everywhere “filling” of the continuum by the first uncountable cardinal without CH

Question

Assuming the negation of CH, let $\omega_1$ be the first uncountable ordinal, $\mathfrak{c}$ be the cardinality of the continuum. Does there exist a map $f: \omega_1 \times [0, 1] \rightarrow \mathfrak{c}$ s.t. for all $t \in \mathfrak{c}$, we have $t \in f(\omega_1 \times \{s\})$ for Lebesgue measure a.e. $s \in [0, 1]$? Obviously this would be true if we assume CH, and assuming the negation of CH this seems impossible intuitively. However, I couldn’t find any simple reason to show no such map can exist. (If it is impossible to prove the existence of such an $f$ without CH, is its existence at least consistent with the negation of CH?)

Answer 1

The existence of such an $f$ is consistent with $\mathsf{ZFC+\neg CH}$. Suppose $\mathfrak{c}=\aleph_2$ and every set of size $\le\aleph_1$ is null (this is consistent with $\mathsf{ZFC}$; it follows, for example, from $\mathsf{2^{\aleph_0}=\aleph_2+MA}$). Fix a bijection $b:[0,1]\rightarrow\omega_2$, and let $f(\omega_1\times\{s\})\supseteq b(s)$. Then for each $\alpha<\omega_2$, there are only $\omega_1$-many reals $s\in[0,1]$ such that $\alpha\not\in f(\omega_1\times\{s\})$.