I'm reading a proof of Steinhauss theorem:
Let $A \subset \mathbb{R}^n$ be a measurable set and $m(A)>0$, where $m$ is the Lebesgue measure. Then, $A-A = \{ x-y: x,y\in A\}$ contains an oper neighbourhood of the origin.
And it uses the following result to prove it:
(1) Let $A \subset \mathbb{R}^n$ be a measurable set then $m(A) = \sup{\{m(F):F\subset A}$ and $F$ closed$\} $
The proof starts taking a compact set $K\subset A$ with $m(K) >0$. This is possible because if $A$ is bounded, we can use (1) to find a close, bounded set, and therefore a compact one.
If $A$ is not bounded, we can guarantee that at least one of these disjoint sets $A \cap (B(0,k) - B(0, k-1))$ for $k \geq 1$ must have a measure greater than $0$ and using (1) again we can guarantee the existence of K.
My problem is in that last paragraph. I think get why it works ($B(0,k) - B(0, k-1)$ is a partition of $\mathbb{R}^n$, but why not using simply $B(0,k)$, of increasing size? Is it important that the sets are disjoint?
The proof doesnt use that $K$ has that "shape". It only uses that $K$ is a compact set contained in $A$. I think I might not fully understand what measurable sets are. Thanks for your help!
I believe you are right. An increasing sequence makes even more sense to me. Here are the details written out:
Note that $$\Bbb{R}^n = \bigcup_{k=1}^\infty B(0,k)$$
Hence, $$A = A \cap \Bbb{R}^n = \bigcup_{k=1}^\infty A \cap B(0,k)$$
Then there is some $k$ such that $m(A \cap B(0,k)) > 0$, otherwise we have $$m(A) \leq \sum_{k=1}^\infty m(A \cap B(0,k))=0$$ which contradicts $m(A) > 0$.
Now, by $(1)$ there is a closed set $F$ with $F \subseteq A \cap B(0,k)$ with $m(F) > 0$. But $A \cap B(0,k)$ is bounded, so $F$ is bounded as well and by Heine-Borel $F$ is compact. Thus $A$ contains the compact set $F$ with $m(F) > 0$.