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 $|A\setminus B|= 0$


1 Answer 1


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 $(t+a_i, t+b_i)$.

  • $\begingroup$ Ah I didn't realize that $(t+A)\setminus (t+B)\subset t+(A\setminus B)$! Thank you very much. $\endgroup$
    – lorenzo
    Sep 2, 2021 at 8:32

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