"Sum" of positive measure set contains an open interval?
Solution 1:
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.
Solution 2:
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.
Solution 3:
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 .