# Cauchy functional equation

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 $(x, y)\in\Bbb R^2$ ?

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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 $$\begin{array} {cl}f(x+y) = f(s+t+y) = f(s) + f(t+y) = f(s) + f(t) + f(y) = f(s+t)+f(y)\\ = f(x) + f(y) \end{array}$$