Show that the following set is Lebesgue measurable

Question

Suppose $A$ is a Lebesgue measurable subset of $\mathbb{R}$ and $ B = \cup_{x\in A}[x-1,x+1]$. Prove that $B$ is lebesgue measurable.

This is exercise 4.17 from Bass's Analysis book http://bass.math.uconn.edu/3rd.pdf

What I'm thinking of doing for this problem is splitting $A$ into countably many pieces as follows: Define $f:\mathbb{R}\to\mathbb{R}, f(x) = x\chi_A(x)$. This function should have countably many jump discontinuities. And taking the preimage of any continuous piece(lets call the preimage $E$) should be a Lebesgue measurable piece of $A$. And furthermore, it is either a point or an interval.(Is that true?). Then $\cup_{x\in E}[x-1,x+1]$ should be lebesgue measurable, and taking the countable union over all such $E$ gives that $B$ is also lebesgue measurable. Is this correct? if so, what else would I need to do to rigorize this argument?

Answer 1

The statement is true even without the measurability assumption on $A$, as proved in the answers by zhw. and Mitchell Spector.

Using measurability however we can write $$B=\left(\bigcup_{x\in A} (x-1,x+1) \right)\cup (A-1)\cup (A+1)$$ so that $B$ is the union of an open set and two translates of $A$, which are all measurable.

Answer 2

Suppose $A\subset \mathbb R$ (not necessarily Lebesgue measurable). Then $\cup _{x\in A} [x-1,x+1]$ is a countable union of intervals (hence is Borel measurable).

Proof: We use the following simple fact: If $\{V_\alpha\}$ is a collection of connected sets such that $\cap_\alpha V_\alpha \ne \emptyset,$ then $\cup_\alpha V_\alpha$ is connected.

In our problem, let's first suppose $A\subset [0,1].$ Then for each $x\in A,$ $[x-1,x+1]$ contains $[0,1].$ Therefore $\cup_{x\in A} [x-1,x+1]$ is connected by the above. And a connected subset of $\mathbb R$ is an interval. Note that we could have used any $[a,a+1]$ in place of $[0,1]$ here.

For the general $A\subset \mathbb R,$ write

$$\cup_{x\in A} [x-1,x+1] = \cup_{n\in Z}(\cup _{x\in A\cap [n,n+1]}\, [x-1,x+1]).$$

This is a countable union of intervals and we're done.

Answer 3

The set $A$ doesn't even have to be measurable for this to be true.

For $x, y\in B,$ define $x\sim y$ iff $[x,y]\subseteq B.$ It's easy to check that this is an equivalence relation on $B.$ (If $y\lt x,$ we're taking $[x,y]$ here to mean the same thing as $[y,x].)$

Every equivalence class is an interval of positive length: If $x\sim y$ and $x\lt z\lt y,$ then $x\sim z,$ so each equivalence class is an interval. Each equivalence class must be an interval of positive (possibly infinite) length, since $x\in B$ implies that there is some $a\in A$ such that $a-1 \le x \le a+1,$ and the entire interval $[a-1,a+1]$ is contained in $B.$

Since the equivalence classes are pairwise disjoint, it follows that there can be only countably many of them. (Each interval contains a rational number, and there are only countably many rationals.)

So $B$ is the union of countably many intervals, and hence is Lebesgue measurable.