Can the unit interval be the disjoint union of countably many "super-dense" parts? I'm curious about this question in the case where $f$ is not necessarily measurable.
I think what it comes down to is this:

Is there an $\varepsilon<1$ and a partition of $[0,1]$ in countably many parts such that every subset of $[0,1]$ with Lebesgue measure at least $\varepsilon$ intersects infinitely many of the parts?

Clearly if the parts are measurable, then this is impossible.
Is it still necessarily impossible when we allow the parts to be wild?
 A: You can do much better. Assuming the axiom of choice, you can partition $\mathbb R$ into $2^{\aleph_0}$ disjoint parts, so that every uncountable Borel set intersects all of the parts.
The construction is a straightforward diagonalization, using the fact that there are only $2^{\aleph_0}$ uncountable closed sets, and each of those sets contains $2^{\aleph_0}$ points. It's the same idea as the construction of a Bernstein set.
Let $\lambda$ be the smallest ordinal of cardinality $2^{\aleph_0}.$ Let $\mathcal A$ be the set of all uncountable closed sets. Let $\mathcal A\times\mathbb R=\{(A_\alpha,t_\alpha):\alpha\lt\lambda\}.$ At step $\alpha,$ choose a point $x_\alpha\in A_\alpha\setminus\{t_\beta:\beta\lt\alpha\}.$ Finally, for each $t\in\mathbb R,$ let $X_t=\{x_\alpha:t_\alpha=t\}.$
The sets $X_t$ are pairwise disjoint, and every uncountable closed set intersects every $X_t.$ It follows that every Lebesgue measurable set of positive measure intersects every $X_t,$ since every set of positive measure contains a closed set of positive measure. In fact, every uncountable analytic set intersects every $X_t.$
