Lebesgue Measure: show that a subset of R is equal to R

Question

If B is a subset of R such that
I) B' has Lebesgue measure zero
II) B is closed under addition

Show that B = R

this is my first course in measure theory. I only know that nonempty close and open subset of R is equal to R.

What do I have from i and ii? And what property of R should I use?

Answer 1

Idea: using the Steinhaus's Theorem:

Let $C$, $D$ be Lebesgue measurable subsets of positive measure. Then the set $C+D$ contains an open interval.

In this case, (i) implies that $B+B$ contains an open interval. But by (ii) $B+B\subset B$, so $B$ contains an open interval $I$. By (i) again, there is $b_0\in I: -b_0\in B$. Now, $I-b_0\subset B$ is a neighborhood of 0 and...

Answer 2

For any real $x$, $B$ and $B - x = \{b - x : b \in B\}$ have full measure so there is some $y \in B \cap B - x$. Hence $y = z - x$ for some $z \in B$. Hence $y, z \in B$ and $x = z - y$. Hence $x \in B$.

This avoids Steinhaus theorem which shows that this is also true if $B$ had just positive measure.