Prove that if $B = \{x-y : x,y \in A\}$, where $A$ is a Borel measurable subset of $R$ with positive measure

Question

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.

Answer 1

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 $(x-\delta,x + \delta) \subset B$

Answer 2

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.