Let $A$ be a a set of real numbers such that $m(A)=1$, where m is lebesgue measure. Prove that there is a subset of $A$, call it $B$, such that $m(B)=1/2$.
I guess that there are many ways to show it - directly from the constracting of the measure, with integrals, from regularity... I would like to see some different ways.
Hint: Define $F:\Bbb R\to\Bbb R$ by $$F(t)=m(A\cap(-\infty,t)).$$Show that $F$ is continuous.
**Answer to the original, with "subgroup" in place of "subset":
Assuming we're talking about additive subgroups: There is no subgroup $A$ with $m(A)=1$ in the first place!
Suppose $m(A)>0$. Then $A\ne\{0\}$, so there exists $x\in A$ with $x>0$. Let $$A_n=A\cap[nx,(n+1)x).$$Then $$A_n=nx+A_0,$$so $m(A_n)=m(A_0)$. Note that $A=\bigcup_n A_n$. So if $m(A_n)=0$ then $m(A)=0$, while if $m(A_0)>0$ then $m(A)=\infty$.