Following the nice answer to Do the Lebesgue-null sets cover "all the sets can naturally be regarded as sort-of-null sets"?, the particular situation that I am especially interested in (which is a kind of "advanced version" of https://math.stackexchange.com/questions/35606/lebesgue-measure-of-the-graph-of-a-function) is:

Let $F$ be the set of bijective Borel-measurable functions $f \colon [0,1] \times [0,1] \to [0,1] \times [0,1]$ that preserve the Lebesgue measure.

Is it the case that for any sequence $(K_n)_{n \in \mathbb{N}}$ of functions $K_n \colon [0,1] \to [0,1]$ and any sequence $(f_n)_{n \in \mathbb{N}}$ in $F$, the set $$ B \ := \ \bigcup_{n \in \mathbb{N}} f_n(\mathrm{graph}\,K_n) $$ does not contain a set of positive Lebesgue measure?

If the answer to the above is

yes: what if we allow the functions $K_n$ to be set-valued functions, with $K_n(x)$ being a Lebesgue-null subset of $[0,1]$ for all $x \in [0,1]$?

[Intuitive motivation: Under suitable conditions, the "disintegration theorem" allows one to regard a measure on a product space $X \times Y$ as an $X$-based random measure on $Y$. From this point of view, one might wish to regard a set whose $Y$-sections are null (under the disintegrated measure) as being itself "effectively null" under the original measure; but if the answers to the above questions are *no*, then we cannot really do this.]