If $E\subseteq\mathbb{R}$ is Lebesgue measurable, show that $$F=\bigcup_{x\in E}[x-1,x+1]$$ is also Lebesgue measurable.
My solution: the 'distance to E' function $$\rho(x) = \inf\limits_{y\in E}|x-y|$$ is continuous and therefore measurable, hence $\rho^{-1}([0,1))$ is measurable. The remaining points in $F$ (those with $\rho(x) = 1$) form a set of measure zero, because they are isolated.
I am looking for alternative solutions and, of course, any corrections to my solution.
I may be missing something, but can you not write $E+[-1,1] = (E+\{-1\}) \cup (E + (-1,1)) \cup (E+\{+1\})$?
Since $E$ is measurable, so are $E+\{-1\}$, $E+\{+1\}$, and since $(-1,1)$ is open, so is $E + (-1,1)$ (hence measurable).