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?
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 $X_t.$