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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?

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  • $\begingroup$ Of course $A$ is measurable: its complement is. $\endgroup$ – Gae. S. Nov 1 '19 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? $\endgroup$ – Marcos G Neil Nov 1 '19 at 18:40
  • $\begingroup$ the $A-A$ problem has $m(A)$ is finite, and I can not do the same thing to this problem $\endgroup$ – Marcos G Neil Nov 1 '19 at 18:41
  • $\begingroup$ @MarcosGNeil But you can do the same to $A \cap [a,b]$ for each interval. $\endgroup$ – N. S. Nov 1 '19 at 19:03
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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 $m(B)=0$.

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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.

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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!

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