# Prove that if a particular function is measurable, then its image is a rect line [duplicate]

I´m really stuck with this problem of my homework. I don´t have any idea, how to begin.

Let $f$ be a function, $f:\mathbb{R}\rightarrow \mathbb{R}$, such that $f(x+y)=f(x)+f(y)$ , $\forall x,y\in\mathbb{R}$. Prove that if f is Lebesgue-measurable, then there exists a $c\in\mathbb R$, such that $f(x)=cx$.

I have the next idea, $\forall \alpha\in \mathbb R$, $f^{-1}((-\infty,\alpha))$, (where $f^{-1}$ denotes the preimage function) is measurable. And somehow I want to end with that the image of such set is $(-\infty,c\alpha)$. Any idea is welcome.

• The condition in your post is often called Cauchy's functional equation and the functions fulfilling this condition are called additive functions. Oct 19, 2013 at 16:47

If $f$ is measurable then its restriction is continuous on some positive measure compact set $K$. Then $f$ is also continuous on $K - K = \{a - b : a, b \in K\}$. This is because if $a_n - b_n$ converges to $c$, then some subsequences $a_{n_k} \rightarrow a$ and $b_{n_k} \rightarrow b$ with $a-b = c$. But $K - K$ contains an interval around zero.
First prove assuming $f$ continuous:
(Prove that $f(\frac{p}{q}\cdot x)=\frac{p}{q}f(x)$ and use the following fact: $\overline{\mathbb{Q}}=\mathbb{R}$)
Then try to prove that a measurable additive function must be bounded in some open neigborhood of $0$ and hence continuous. Here are the details.