A possible measure-theoretic pathology

Question

Let $S$ be a nonempty closed subset of the open unit square $(0,1)^2 = X \times Y$ that has the following "shadow property":

For any aligned open square $C = A \times B$ that intersects $S$, the projections of $S \cap C$ on $A$ and on $B$ have positive Lebesgue measure.

(Examples: any set that is the closure of its interior; any curve that doesn't contain a vertical or horizontal line segment.)

Let $|\cdot|$ denote Lebesgue measure, and for any $W \subset X$, let $N(W)$ be the set of all $y \in Y$ such that $(x,y) \in S$ for some $x \in W$.

Can there be, for some $S$, a set $W \subset X$ with $|W| = 1$ but $|N(W)| = 0$?

Answer 1

Let $N$ be a dense null set and $S = (([0,1] \setminus N) \times N) \cup (N \times [0,1])$ and let $W = [0,1] \setminus N$. Then $N(W) = N$ so $W$ is as needed.

Let $A$ and $B$ be open intervals. Then $S \cap A \times B$ contains $(A \cap N) \times B$ so, as $N$ is dense, $S \cap A \times B$ projects to $Y$ as $B$. Likewise, $A \times B$ contains $(A \setminus N) \times (B \cap N)$ so projects to $X$ as $A \setminus N$.

Answer 2

Suppose that $S$ is the graph of a continuous, strictly increasing function $\psi$. Then $S(W) = \psi(W)$, and the question asks if there is $\psi$ and $W$ such that $|W| = 1$, but $|\psi(W)| = 0$. The answer is affirmative.

Let $\mu$ be a singular probability measure on $(0, 1)$ with no atoms and full support, and let $\psi(x) = \mu((0, x))$ denote its distribution function. Then $\mu(A) = |\psi(A)|$ for every $A = (0, x)$, and hence for every Borel $A$. Let $V$ be a Borel set such that $|V| = 0$ and $\mu(V) = 1$. Then $W = (0, 1) \setminus V$ satisfies $|W| = 1$ and $|\psi(W)| = \mu(W) = 0$, as desired.