# Prove $A+A=\mathbb{R}$ in which the measure of the complement of $A$ is zero

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?

• Of course $A$ is measurable: its complement is. – Gae. S. Nov 1 '19 at 18:37
• 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? – Marcos G Neil Nov 1 '19 at 18:40
• the $A-A$ problem has $m(A)$ is finite, and I can not do the same thing to this problem – Marcos G Neil Nov 1 '19 at 18:41
• @MarcosGNeil But you can do the same to $A \cap [a,b]$ for each interval. – N. S. Nov 1 '19 at 19:03

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

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

while

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

Thus,

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

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