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Let $B \subset \mathbb{R}$ with $m(\mathbb{R}\setminus B)=0$. Show $B+B=\mathbb{R}$ , where $m$ is a Lebesgue measure.

I have been trying to modify the solution posted by @tristan (cf this solution) to show instead $B+B=\mathbb{R}$.

How does one define $B$ so that $B+B=\mathbb{R}$?

user565684
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    "How does one define $B$ so that $B+B=\mathbb{R}$?" -- I'm not sure what you're asking by this. It feels like you want a specific example of such a $B$, but bear in mind that this proof has to apply for any set $B$ (which is a subset of the reals and that the reals have measure $0$ when removing that set). One particular example won't cut it. – PrincessEev Oct 29 '19 at 05:33
  • Yes, I meant for it to apply for any set $B$ – user565684 Oct 29 '19 at 06:08

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Let $x$ be any real number. Then $(x-B^{c}) \cup B^{c}$ has measure $0$ because $m(x-B^{c})=m(B^{c})$. Hence there must be at least one real number $t$ in the complement of $(x-B^{c}) \cup B^{c}$. This means $t \in (x-B)\cap B$. Hence we can write $t=x-b_1=b_2$ for some $b_1,b_2 \in B$. This gives $x=b_1+b_2 \in B+B$.

Kavi Rama Murthy
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