# “Sum” of positive measure set contains an open interval?

So this homework question is in the context of $\mathbb{R}$ only, and we are using Lebesgue measure.

The sum $A+B$ is defined to be $A+B=\{a+b|a\in A,b\in B\}$. The question is: If $m(A),m(B)>0$, is it necessarily the case that $A+B$ contains an open interval?

This question is quite annoying, since it is quite difficult to calculate the sum even after you construct an example. However, it seems like no matter how I constructed a set of positive measure, there is a region that have a lot of point in it (yet this region is not necessarily dense). This suggest that when summed up, the holes should be covered up somehow and thus there would be an open interval, and so the answer is yes. Sorry for a rather vague idea, but I'm just not sure what to do with this problem. I have a feeling that somehow topology would have a way to capture precisely the idea above about a region with a lot of point.

(also, I think my middle school teacher mentioned this once, and I remembered the answer to be yes; but that is quite a while ago though)

Thank you. Any help will be appreciated.

EDIT2: So Camilo's answer is gone. I am looking at the convolution method right now. Because the convolution is continuous, if it is nonzero I am done. But then there is no guarantee that the convolution is nonzero even if $x\in A+B$. I still need help with this direction. Can someone give me some hints?

EDIT: Camilo's answer below essentially suggest the method of getting 2 open interval from regularity theorem, one cover a subset of A and one cover a subset of B such that the sum of the measure of these two subset is bigger than the measure of the larger interval. user99680's answer point to a wiki which show a method that essentially simply acquire an open set. After some thinking, I believe that these method simply can't be made to work. I constructed this example in the hope that it would convince you that it is the case:

Let $I_{0},I_{1},\ldots$ be disjoint open interval such that the distance between any 2 of them is nonzero and $m(I_{n})=9^{-n}$ and all of them are inside $[0,2]$. Let $J_{0},J_{1},\ldots$ be disjoint open interval such that the distance between any 2 of them is at least $3$ and $m(J_{3n})=m(J_{3n+1})=m(J_{3n+2})=\frac{m(I_{n})}{3}$. Let $O=\bigcup I_{n}$ and $H=\bigcup J_{n}$. Construct $A\subset O$ such that $m(A\bigcap I_{n})=\frac{3m(I_{n})}{4}$. Construct $B\subset H$ such that $m(B\bigcap J_{n})=\frac{3m(J_{n})}{4}$. Now if you apply Camilo's method, it is possible that your $I$ would be one of the $I_{n}$ above, and your $J$ is one of the $J_{m}$ above. No matter which $I_{n},J_{m}$ you have, the argument does not work. Rewrite $m=3q+r$ where $0\leq r\leq 2$. If $q<n$ then $m(J_{m})>m(I_{n})$ so the longer interval have length $\frac{m(I_{q})}{3}$; $m(A\bigcap I_{n})$ is at most $\frac{3m(I_{q+1})}{4}=\frac{m(I_{q})}{12}$ and $m(B\bigcap J_{m})=\frac{3m(J_{m})}{4}=\frac{m(I_{q})}{4}$. Thus $m(A\bigcap I_{n})+m(B\bigcap J_{m})$ is no larger than the larger interval. If $n\geq q$ then $m(I_{n})>m(J_{m})$ so the larger interval is of length $m(I_{n})$; $m(A\bigcap I_{n})=\frac{3m(I_{n})}{4}$ and $m(B\bigcap J_{m})$ is at most $\frac{3m(J_{3n})}{4}=\frac{m(I_{n})}{4}$. Once again $m(A\bigcap I_{n})+m(B\bigcap J_{m})$ is no larger than the larger interval. Hence the Camilo's argument does not work. user99680's method does not work either, since you get $O$ and $H$ which is even bigger than any interval.

• Well, I couldn't come up with an answer, in this page the problem done: blog.jpolak.org/?p=177. – Camilo Arosemena-Serrato Dec 8 '13 at 3:09
• Ah thanks for the information. This seems like a very interesting method, but it also seems way too advanced. I have not learnt any of these theorem yet. I was hoping for some elementary methods, at least for this homework. I will look further into that method though. – Gina Dec 8 '13 at 13:36
• possible duplicate of Steinhaus theorem (sums version) – ronno Dec 15 '13 at 11:12
• Answered in math.stackexchange.com/questions/86209/…, although the question is a special case. – ronno Dec 15 '13 at 11:12
• The upvoted & accepted Answer to the Question noted by @ronno is exactly the result asked for. – hardmath Dec 15 '13 at 13:27

Hint: Use the fact that Lebesgue measure is regular to show that for any set $X$ of positive measure, there's an $\varepsilon_0$ such that for any $\varepsilon<\varepsilon_0$ there's an interval $I$ of length $2\varepsilon$ such that $X\cap I$ has measure greater than $\varepsilon$.

Then use translation and reflection invariance of Lebesgue measure to reduce your problem to the Steinhaus theorem.

• Can you give some more hints please? I still can't progress on this. On both part in fact. I can't figure out how that work for all $\epsilon<\epsilon_{0}$. Also, I don't know how that reduce back to Steinhaus problem, since the 2 sets can still be very radically different even after a finite number of translation and reflection of their subsets (just a hunch). – Gina Dec 8 '13 at 13:27
• @Gina: Well, it takes a little more than just regularity. The statement I cited follows from Lebesgue density theorem, which does use the geometry of ${\bf R}$ quite a bit. The second part, however, is not that hard. Think about it for a while and ask again if you still have trouble. – tomasz Dec 8 '13 at 16:57
• Oh I can solve the problem if I manage to get the first part, but not by reducing it to Steinhaus problem. If I get the first part, then I can find 2 interval of the same length $(a-r,a+r)$ and $(b-r,b+r)$ such that $m(A\bigcap(a-r,a+r)),m(B\bigcap(b-r,b+r))>r$. Then $\chi_{A}\ast\chi_{B}(a+b)>0$ and then use continuity of convolution of compact support function to obtain an interval. But this method is not the same as reducing to Steinhaus problem, so I wonder how you would do it or is this already is similar to what you meant when you said so. – Gina Dec 8 '13 at 18:33
• As for density theorem, unfortunately I don't study that one, and I guess I can't put the whole proof into the homework. There was a mention of the density theorem in the comment above, but I am looking to something more elementary. – Gina Dec 8 '13 at 18:35
• @Gina: If you have two intervals $I$, $I+r$ covered in more than half measure-wise by $A,-B$ respectively, then $-(B+r)\cap A$ has positive measure (it contains a subset of $I$ of positive measure), so by Steinhaus' theorem $(-(B+r)\cap A)-(-(B+r)\cap A)$ has nonempty interior. But then of course $A-(-(B+r))=A+(B+r)$ and so does $A+B$. – tomasz Dec 9 '13 at 14:14

This method is rather elementary. It basically use what can be considered a generalization of pigeon hole principle.

First reduce the problem to finite measure set $A,B$.

Then prove this version of pigeon hole principle: if $P_{1},P_{2},\ldots$ is a countable disjoint partition of finite positive measure set $S\supset A$ then for any $\epsilon<\frac{m(A)}{m(S)}$ there exist an $P_{n}$ such that $\epsilon<\frac{m(A\bigcap P_{n})}{m(P_{n})}$. To prove this, you pretty much do it the same way you would prove the normal pigeon hole principle.

Once you got it, then using regularity, you can have a set $U\supset A$ and $V\supset B$ such that $\frac{1}{2}<\frac{m(A)}{m(U)},\frac{m(B)}{m(V)}$. Using $U$ being a disjoint countable union of open interval, apply the above "pigeon hole" to get an interval $I$ where $\frac{1}{2}<\frac{m(A\bigcap I)}{m(I)}$. Now for $V$ it is slightly more complicated. First partition into disjoint open interval. Now each interval can be partitioned into a countable number of interval in which each have length $\frac{m(I)}{n}$ where $n\in\mathbb{N}$. Countable times countable is still countable, so once we do that for all open interval, we just partitioned $V$ into a countable number of interval wherein each is of length of the form $\frac{m(I)}{n}$. One of them, $J$ would have $\frac{1}{2}<\frac{m(B\bigcap J)}{m(J)}$ and $m(J)\times n=m(I)$. Now we partition $I$ into $n$ interval of length same as $J$, and using "pigeon hole" above, one of them, $L$ would have $\frac{1}{2}<\frac{m(A\bigcap L)}{m(L)}$. Now we have $J,L$ having the same length, and $A,B$ each occupies more than half of them. The rest is trivial.

See Steinhaus theorem: http://en.wikipedia.org/wiki/Steinhaus_theorem , where this is proved. It is not exactly what you're looking for, but it will help get you started .

• I don't think the method for that proof can be made to work at all. See my edited question for a possible reason. That method pretty much require knowing that $K$ is fully inside $U$. In the above example, the $K$ of one set cannot fit into the $U$ of the other set. I think we need a different approach here. – Gina Dec 7 '13 at 5:58