Is $\bigcup_{x \in A} [x - 1, x + 1]$ Lebesgue measurable, where $A$ is a Lebesgue measurable subset of $\mathbb{R}$? Suppose $A$ is a Lebesgue measurable subset of $\mathbb{R}$ and $$B = \bigcup_{x \in A} [x - 1, x + 1].$$Is $B$ Lebesgue measurable?
 A: We have:
$$B = \bigcup_{x \in A} (x-1, x+1) \cup \bigcup_{x \in A} \{x-1\} \cup \bigcup_{x \in A} \{x +1\} = O \cup (A - 1) \cup (A + 1)$$
$O$ is open so Lebesgue measurable, and $A \pm 1$ are Lebesgue measurable since so is $A$.
A: The answer is yes, even for $A$ being an arbitrary index sets. The following is from Problem 2.J. of General Topology by Kelley:

A proper interval is defined to be an open/closed/half-open interval with different endpoints. Let $\mathscr{C}$ be an arbitrary family of proper intervals. Then there is a countable subset $\mathscr{B}$ of $\mathscr{C}$, such that $\bigcup \mathscr{B} = \bigcup\mathscr{C}$.

Put $C = \bigcup \mathscr{C}$. First, we prove that all but at most countably many points in $C$ are interior points of some $I\in \mathscr{C}$. If $x\in C$ is not an interior point of any $I\in \mathscr{C}$, then it's an end point of some (half-)closed interval in $\mathscr{C}$. Suppose we have uncountably many such $x$. WLOG, suppose that uncountably many of them are left end points. Let $S$ be the set of them.
To each $s\in S$ there corresponds a proper interval $[s,t) \subset I$ for some $I\in \mathscr{C}$, such that $[s,t)\cap S = \{s\}$ (because other points in $S$ are not interior points of $I$). Hence these proper intervals $[s,t)$ are pairwise disjoint (otherwise, say $[s_1,t_1)$ intersects $[s_2,t_2)$ and $s_1 < s_2$, then $s_2 \in [s_1,t_1)$). But $\mathbb{R}$ allows at most countably many pairwise disjoint proper intervals (because it's separable), which is a contradiction.
Let $C'$ be the set of points $x\in C$ such that $x\in I^\mathrm{o}$ for some $I\in \mathscr{C}$, the previous discussion shows that $C-C'$ is at most countable. Note that $\{I^\mathrm{o}:I\in \mathscr{C}\}$ is an open cover of $C'$. Since $\mathbb{R}$ is secound-countable, this open cover admits a countable subcover. Let $\mathscr{B}$ be the set of proper intervals corresponding to this subcover, together with one element of $\mathscr{C}$ for each $x\in C-C'$ that contains $x$. Then $\mathscr{B}$ is countable, and $\bigcup \mathscr{B} = C$.
