Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there $U\subset \Bbb R^2$ with Lebesgue measure $0$ such that

$$f(x+y)=f(x)+f(y)$$ for all $(x, y)\in U$ implies $f(x+y)=f(x)+f(y)$ for all $(x, y)\in\Bbb R^2$ ?

share|cite|improve this question

Let $E$ be a subset of $\mathbb R$ with Lebesgue measure $0$ such that $E + E = \mathbb R$ (e.g. the union of integer translates of the Cantor set). Let $U = E \times {\mathbb R}$, which has two-dimensional Lebesgue measure $0$. Suppose $f(x+y) = f(x) + f(y)$ for $(x,y) \in U$. Given any $(x,y)$, we have $x = s + t$ for some $s,t \in E$, and $$\begin{array} {cl}f(x+y) = f(s+t+y) = f(s) + f(t+y) = f(s) + f(t) + f(y) = f(s+t)+f(y)\\ = f(x) + f(y) \end{array}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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