# if $A\subset\mathbb{R}$ is Lebesgue measurable and $t\in\mathbb{R}$, then $t+A$ is Lebesgue measurable

I have proved part of the following statement and I would like to have an hint about how I could finish up my proof, thanks.

"Prove that if $$A\subset\mathbb{R}$$ is Lebesgue measurable and $$t\in\mathbb{R}$$, then $$t+A$$ is Lebesgue measurable"

What I have done:

If $$A$$ is a Borel set then $$t+A$$ is also a Borel set and $$|(t+A)\setminus (t+A)|=|\emptyset|=0$$ so $$t+A$$ is Lebesgue measurable too, by definition.

In the following we suppose that $$A$$ is not a Borel set.

We first consider the case $$|A|<\infty$$: then by definition of Lebesgue measurable set there exists a Borel set $$B\subset A$$ such that $$|A\setminus B|=0$$. Now let $$t\in\mathbb{R}$$: then $$t+B$$ is a Borel set, $$t+B\subset t+A$$ and $$|t+B|=|B|\leq |A|<\infty$$ (outer measure is translation invariant) so $$|(t+A)\setminus (t+B)|=|t+A|-|t+B|=|A|-|B|=|A\setminus B|=0$$ so $$t+A$$ is Lebesgue measurable too.

The problematic part

I tried to prove the case $$|A|=\infty$$ in two ways: by showing that $$\bigcup_{k=1}^{\infty}(t+A_k)=\bigcup_{k=1}^{\infty}(t+A\cap[-k,k])$$ is such that $$|A\setminus \bigcup_{k=1}^{\infty}(t+A_k)|=0$$ but then I realized that $$A_k$$ is not necessarily Borel so this fails and then I also tried by contradiction, by assuming that $$|(t+A)\setminus (t+B)|>0$$ (where $$B\subset A$$ is the Borel set such that $$|A\setminus B|=0$$) and from this I got that there must be an uncountable number of elements in $$|A\setminus B|$$ but then I realized that this doesn't necessarily mean that $$|A\setminus B|>0$$ as there exist uncountable sets which have Lebesgue measure $$0$$ (e.g the Cantor set) so I am stuck.

DEF. (Lebesgue measurable set): A set $$A\subset\mathbb{R}$$ is called Lebesgue measurable if there exists a Borel set $$B\subset A$$ such that $$|A\setminus B|= 0$$

There exists a Borel set $$B$$ contained in $$A$$ such that $$|A \setminus B|=0$$. Now $$C=t+B$$ is a Borel set contained in $$t+A$$ and $$(t+A)\setminus (t+B)$$ is contained in $$t+ (A\setminus B)$$ so $$|(t+A)\setminus (t+B)|=0$$ .
I have used the fact that if $$|E|=0$$ then $$|t+E|=0$$ for any real number $$t$$. For the proof cover $$E$$ by open intervals $$(a_i,b_i)$$ with $$\sum (b_i-a_i) <\epsilon$$ and consider the intervals $$(t+a_i, t+b_i)$$.
• Ah I didn't realize that $(t+A)\setminus (t+B)\subset t+(A\setminus B)$! Thank you very much. Sep 2, 2021 at 8:32