We know that for a Lebesgue measurable set $S \subseteq \mathbb{R}$, any of its translates $x+S$ for some $x \in \mathbb{R}$ is Lebesgue measurable with $m(x+S) = m(S)$. I am wondering about properties of higher-dimensional sets that follow a similar reasoning.
Set $S = [0, 1]^2 = [0, 1] \times [0, 1]$. Denote $S_y = \{x: (x, y) \in S\}$ (the part of $S$ at height $y$). My question is what happens if I translate some of these $S_y$'s? Will I still be left with a measurable set, and if so does the measure of this new set agree with the measure of the original set?
To be explicit, I mean what happens if we take $$T = \left(S \setminus\bigcup_{y}(S_y \times\{y\})\right) \cup \bigcup_{y}((S_y + t_y) \times \{y\}),$$ where the union is taking over some set of $y$'s, will this set be measurable? Of course if we take the union over a set of measure $0$ then both questions have an answer of yes. Additionally, if we translate the whole set $S$, then of course both answers still have a positive answer of yes. However how about in other cases? It seems to me that one could produce a non-measurable set this way. Are there any examples anyone knows of?