$h:\mathbb{R} \rightarrow 2\pi\mathbb{Z}$ and $h(x+y)=h(x)+h(y)$ then $h \equiv 0$.

Question

I don't understand Daniel's proof where he said $h:\mathbb{R} \rightarrow 2\pi\mathbb{Z}$ and $h(x+y)=h(x)+h(y)$ so $h \equiv 0$ cause I don't see any problem putting $h(1)=2n\pi$ for some $n \in \mathbb{Z}$. Can anyone elaborate this?

Proving that $f$ is measurable with $f(x+y)= f(x)+f(y)$ then $f(x) =Ax$ for some $A\in\Bbb R$?

Answer 1

Suppose $h(1)=2\pi n$. For any positve interger $k$ we have $kh(\frac 1 k)=h(1)$ by additivity so $h(\frac 1 k) =2 \pi \frac n k$. This must be $2 \pi m$ for some integer $m$. This implies that every integer $k$ divides $n$, so $n$ must be $0$. Thus $h(1)=0$. A similar argument shows that $h(x)=0$ for every $x$.