In Stein's textbook Chapter 2, page 75-76. We read:
"...even the assumption that $f$ is measurable on $\Bbb{R}^d$, it is not necessarily true that the slice $f^y$ is measurable on $\Bbb{R}^d$ for each $y$; nor does the corresponding assertion necessarily hold for a measurable set $E$: i.e. the slice $E^y$ may not be measurable for each $y$. An easy example arises in $\Bbb{R}^2$ by placing a one-dimensional non-measurable set on the x-axis; the set $E$ in $\Bbb{R}^2$ has measure zero, but $E^y$ is not measurable for $y = 0$. What saves us is that, nevertheless, measurability holds for almost all slices."
All good. My problem is that in the book we assumed that the Lebesgue measure is complete. Yet here the authors provide an example of a non-measurable subset of a zero-measure set? Does anyone know this counter-example (for the set with zero-measure) that they are talking about? And also can anyone please explain what am I not understanding regarding this kind of subset w.r.t. the Lebesgue measure?
That is not what happened here.
In $\mathbb{R}^2$, the non-measurable set $A$ on the $x$-axis takes the form $E := A \times \{0\}$, and hence has measure zero.
The set $E^y = \{ x \mid (x,y) \in E \} = A$ is a subset of $\mathbb{R}$, and characteristics tied to its measure are tied to the measure in $\mathbb{R}$.
That is to say, $E^y$ (a non-measurable set in $\mathbb{R}$) is not a subset of $E$ (a zero-measure set in $\mathbb{R}^2$).