# 1. Measurable sets : uncountable union, 2. null set is disjoint from translated rationals

I am doing exercises in Real analysis for Graduate students by Richard F. Bass and I am stuck.

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

I think it is immediate if $A$ is countable. However, for uncountable set $A$. I was given a hint to work on open interval, then come to closed interval later. Let $$B' = \cup_{x \in A} (x-1,x+1).$$ Then $B$ is open set in $\mathbb{R}$, so $B$ can be written as a countable union of disjoints open intervals, and hence measurable. I see that $B = B' \cup (\cup_{x \in A}\{x-1, x+1\})$. But if $A$ is uncountable, I stuck with $$\cup_{x \in A}\{x-1, x+1\}.$$ Any suggestion to move on from this ?

1. If $A \subseteq \mathbb{R}$ with zero Lebesgue measure ($m(A) =0$), then $$A \cap (c + \mathbb{Q}) = \phi$$ for some $c \in \mathbb{R}.$ I was guided to look at $\cup_{q \in \mathbb{Q}}(q+A)$. But I find nothing from there except $m(q+A) = 0$ for all $q \in \mathbb{Q}.$

For the first task, e.g. use that your remaining set $\bigcup_{x\in A} \{x+1,x-1\}$ is the union of two translations of $A$, and that the translation of a measurable set is measurable.

For the second task, $\bigcup_{q\in\mathbb{Q}} (q+A)$ would equal all of $\mathbb{R}$ if $A\cap (c+\mathbb{Q}) \neq \emptyset$ for all $c\in\mathbb{R}$: Given $c\in \mathbb{R}$, there would always be $q\in \mathbb{Q}$ and $x\in A$ so that $c+q = x$, i.e. $c = -q+x \in q+A$. But that union is a null set as a countable union of null sets.

• We have $A+1 = \{ x+1 \, |\, x\in A\} = \bigcup_{x\in A} \{x+1\}$ and $A-1 = \{ x-1 \, |\, x\in A\} = \bigcup_{x\in A} \{ x-1\}$. These are the definitions of the translations of $A$ by $1$ and $-1$ respectively. Then clearly $\bigcup_{x\in A} \{x+1,x-1\} = \bigcup_{x\in A} \{x+1\} \cup \bigcup_{x\in A} \{x-1\} = (A+1) \cup (A-1)$. – user35359 Sep 15 '15 at 16:05
• Note that I was talking about the set $\textbf{remaining}$ after subtracting the obviously open part, as in Both Htob's approach. – user35359 Sep 15 '15 at 20:36
• Um, I can't see anything wrong with my post as it is now though. – user35359 Sep 15 '15 at 21:24

Regarding the first question: Curiously, $B$ is Lebesgue measurable, in fact Borel measurable, for any set $A$, measurable or not. We only need to show that $B\cap[n,n+1]$ is measurable. But if you consider the various possibilities you see that $B\cap[n,n+1]$ is at worst the union of two intervals.

1. In fact $B$ is the union of interval $[x-1, x+1]$ centered at $x\in A$ with length $2$. So for any $x,y\in A$ that $|x-y|<2$, $$\:[x-1, x+1]\cup [y-1, y+1]=[x-1, y+1]$$ is a connected interval. As the result, $B$ is the union of disjoint intervals that the number of intervals is at most countable, i.e. $$B=\bigcup_{n\in S}I_n,\quad I_n\in\{(a,b),[a,b),(a,b],[a,b]\},\:I_n\cap I_m=\varnothing,\:\forall n,m\in \Bbb{N}, \:n\ne m$$ and S is either finite or countable infinite. Hence $B$ is Lebsgue measurable.

2. Suppose for all $c\in \Bbb{R}$, $A \cap (c + \mathbb{Q}) \ne \varnothing$. We prove that $$\bigcup_{q \in \mathbb{Q}}(q+A)=\Bbb{R}$$ Let $r\in A \cap (c + \mathbb{Q})$. Then $\exists q'\in \mathbb{Q},\:r=c+q'\:$ and $\:\exists a\in A, \:r=a$. So $$c=-q'+a\in \bigcup_{q \in \mathbb{Q}}(q+A)$$ which means $\Bbb{R}\subset \bigcup_{q \in \mathbb{Q}}(q+A)$. Since $\bigcup_{q \in \mathbb{Q}}(q+A)\subset \Bbb{R}$, we have $$\bigcup_{q \in \mathbb{Q}}(q+A)=\Bbb{R}$$ But since for any $q\in\mathbb{Q},\: m(q+A)=m(A)=0$, $$m(\bigcup_{q \in \mathbb{Q}}(q+A))\leqslant\sum_{q\in\mathbb{Q}}m(q+A)=0$$ But $m(\Bbb{R})=\infty$, which is contradiction.