Non-measurable sets from a well-ordering of the continuum

Prove that there exists a non-measurable set $E \subset [0,1] \times [0,1]$ with the property that it intersects every vertical line at exactly one point.

By Example 1.4.13, there is a well-ordered set $W$ and a bijection $j:[0,1] \rightarrow W$ such that for any $x \in [0,1], j(x)$ has at most a countable number of predecessors. Let $Q$ be the set of points $(x, y)$ in $[0,1] \times [0,1]$ such that $j(x)$ precedes $j(y)$, and let $f=\chi_{Q}$. Prove that (8.9) is not true for $(X, \mathfrak {M}, \mu )=(Y, \mathfrak {N}, \lambda )=[0,1]$ with Lebesgue measure. Explain why it does not contradict Theorem 8.2.1.

Theorem 8.2.1, with formulas (8.8) and (8.9): Let $(X, \mathfrak {M}, \mu )$ and $(Y, \mathfrak {N}, \lambda )$ be $\sigma$-finite measure spaces and let $Q \in \mathfrak {M} \otimes \mathfrak {N}$. If $\varphi (x)=\lambda (Q_{x})$ and $\psi (y)=\mu (Q^{y})$ (formula (8.8)), where $Q_{x}=\left\{ y \in Y: (x, y) \in Q\right\}$ and $Q^{y}=\left\{ x \in X: (x, y) \in Q\right\}$, then $\varphi$ is $\mathfrak {M}$-measurable, $\psi$ is $\mathfrak {N}$-measurable and

which is equivalent to $\int_{X} d\mu (x) \int_{Y} \chi_{Q}(x, y) \, d\lambda (y)=\int_{Y} d\lambda (y) \int_{X} \chi_{Q}(x, y) \, d\mu (x)$.