Let $A\subset \mathbb{R}$ such that $m(\mathbb{R}\smallsetminus A)=0$. Show that $A+A=\mathbb{R} $, where $$A+A=\{a+b\mid a,b\in A\}$$.

This is a question in the past qualifying exam in my university. I do not know where to approach. I encountered a similar problem that if $A$ is measurable and $m(A)>0$, then $A-A$ contains an interval, but I used $m(A)$ finite to do this problem. Can you help?

  • $\begingroup$ Of course $A$ is measurable: its complement is. $\endgroup$
    – user239203
    Nov 1, 2019 at 18:37
  • $\begingroup$ right, I forgot R is complete. I have tried the same method of the $A-A$ problem, but it does not work that way, can you give me a hint? $\endgroup$ Nov 1, 2019 at 18:40
  • $\begingroup$ the $A-A$ problem has $m(A)$ is finite, and I can not do the same thing to this problem $\endgroup$ Nov 1, 2019 at 18:41
  • $\begingroup$ @MarcosGNeil But you can do the same to $A \cap [a,b]$ for each interval. $\endgroup$
    – N. S.
    Nov 1, 2019 at 19:03

3 Answers 3


Suppose not. Then there exists $r\in \mathbb{R}$ s.t. $r-a\notin A$ for any $a\in A$. Let $B:=\{r-a: a\in A\}$. By translation invariance of the Lebesgue measure $m(B)>0$. However, $B\subset \mathbb{R}\setminus A$ and so $m(B)=0$.


Assume a point $p$ does not exist in $A + A$.

That means that $\forall x, (x \in A)\rightarrow((p-x) \not \in A)$.

Define $A_p = \{p-x| x\in A \}$ . We have $A_{p} \in \bar{A}$ .

However, $f(x) = p-x$ is a measure preserving transformation, so $A_p$ is of the same measure as $A$. Therefore the complement of $A$ would have measure greater than or equal to $A$.

If it had measure $0$, then the union of $A$ and its complement would be of measure $0$. This cannot happen.


Against the thesis, let $\ q\in\mathbb R\setminus(A\times A).\ $ Let $\ p:= \frac q2.\ $ Define,

$$ K\ :=\ [p-1;p]\cap\mathbb R\qquad\mbox{and}\qquad L\ := [p;p+1] $$ Furthermore, Let $\forall_{x\in\mathbb R}\ f(x):= q-x\ $ and

$$ K'\ :=\ f(K) $$

so that

$$ K'\ \subseteq [p;p+1]\qquad\mbox{and}\qquad K'\cap L=\emptyset. $$

Clearly, all three sets $\ K\ K'\ L\ $ are measurable, and $\ m(K')+m(L)\ \le\ 1.\ $ But $\ m(K')=m(K),\ $ hence

$$ m(K\cup L)\ =\ m(K)+m(L)\ \le\ 1 $$


$$ \mathbb R\cap[p-1;p+1]\ =\ K\cup L $$


$$ m((\mathbb R\cap[p-1;p+1])\ \setminus\ (K\cup L))\,\ \ge\,\ 1 $$

This proves that $\ m(\mathbb R\setminus A)\ \ge\ 1,\ $ in a contradiction to the assumption of the theorem. Great!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.