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.

  • $\begingroup$ The condition in your post is often called Cauchy's functional equation and the functions fulfilling this condition are called additive functions. $\endgroup$ – Martin Sleziak Oct 19 '13 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.