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?)



Almost everywhere “filling” of the continuum by the first uncountable cardinal without CH
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 [$\mathsf{2^{\aleph\_0}=\aleph\_2+MA}$](https://en.wikipedia.org/wiki/Martin%27s_axiom))*. 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 $\alpha\not\in f(\omega\_1\times\{s\})$.


