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Suppose that $m$ is Lebesgue measure, and $A$ is a Borel measurable subset of $R$ with $m(A) > 0$. Prove that if $B = \{x - y : x,y \in A\}$, then $B$ contains a non-empty open interval centered at the origin (Steinhaus theorem).

My attempt at a solution:

I have two ideas for this proof. My first idea was to show that it was true for intervals, and then generalize to Borel measurable sets using the $\pi-\lambda$ theorem, but it doesn't seem that the set of all subsets of $R$ such that $B$ (defined as above) contains an open interval is a $\lambda$-system.

The second idea, which was a hint that was given to me, was to consider the function $f(x) = m((x+A)\cap A)$. If this function could be shown to be continuous, then we could consider $f(0) = m(A) > 0$. I don't really know where this gets us, though.

poppy3345
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  • Did you mean $x-y$ in the definition of $B$? I suspect so, since the result you want says $0\in B$. Assuming yes: If you can show $f$ is continuous then $f(x)>0$ for all $x$ in some interval. And it's easy to see that $(x+A)\cap A\ne\emptyset$ implies $x\in B$. – David C. Ullrich Jul 08 '15 at 00:51
  • So, we show that $0 \in B$, then say that since $f$ is continuous, for some interval around $0$, $f(x) > 0$, therefore, $x \in B$ for all $x$ in that interval? @DavidC.Ullrich – poppy3345 Jul 08 '15 at 01:15
  • Also, where do we use that $A$ is Borel measurable here? In the proof of continuity? – poppy3345 Jul 08 '15 at 01:19
  • Yes. Yes. (You should note the answer Conrada gave - that's what I was going to say before I saw the hint you'd been given. After you finish this exercise, the next one is to convince yourself that the two solutions are close to the same.) – David C. Ullrich Jul 08 '15 at 02:04
  • OH btw in both solutions we should really begin by saying wlog $m(A)<\infty$. – David C. Ullrich Jul 08 '15 at 02:08
  • @DavidC.Ullrich thank you! I will confirm Conrado's answer, as well - the book I am using has this as an exercise before covering convolutions, so I wanted to see a way to do it without them, but I will confirm it for myself with convolutions as well. – poppy3345 Jul 08 '15 at 02:55

2 Answers2

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Clearly for any $x$ and Lebesgue measurable set $A$, $m(x+A)=m(A)$.

If $x\notin B=A-A$, then $$(A+x)\cap A=\varnothing\hspace{4 mm} \text{i.e.}\hspace{4 mm} m((A+x)\cup A)=2m(A)$$ We can suppose that $m(A)>0$ and $A$ compact, as we can limit it to a compact subset of finite measure.

Suppose it's not true. Then there is a sequence $x_n\to0$ such that $x_n\notin A-A$.

On one hand $m((A+x_n)\cup A)=2m(A)$. On the other hand, $m((A+x_n)\cup A)\to m(A)$ for $A$ is compact and there is finite cover of intervals with length approaching $m(A)$. So $2m(A)\to m(A)$ is clearly a contradiction.

Eugene Zhang
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Let $f = 1_A$ $g = 1_A$ take $h=f * g $

note that $$\|f*g\|_1 = \|f\|_1 \|g\|_1.$$

Since $\text{supp} h \subset A+A = B $, $h$ is continuous and $\|h\|_1>0$ there is a $x \in \Bbb{R}$ such that $h(x)>0$. Use the continuity of $h$ to conclude that $h(y)>0$ for every $y \in (x - \delta, x + \delta)$. So $(x-\delta,x + \delta) \subset B$

Conrado Costa
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  • It's true that $||fg||\infty=||f||\infty||g||_1$ here, but that depends on exactly what $f$ and $g$ are. Some might feel it was in better taste to use the fact that $\int fg=\int f\int g$, since that holds for any $f$ and $g$ (and hence it's easier to see why it's true. For me anyway). – David C. Ullrich Jul 08 '15 at 02:07
  • Lemme retract that: I don't actually believe that $||f*g||\infty=||f||\infty||g||_1$. – David C. Ullrich Jul 08 '15 at 02:13