Hartman's theorem (Section 3)

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Source. Sections 1 and 3, printed pp. 59--61 (PDF pp. 1--3). De Bruijn attributes the original result to S. Hartman, A remark about Cauchy's equation, Colloquium Mathematicum 8 (1961), 77--79.

Statement. Let $S\subseteq\mathbb R$ have Lebesgue measure zero. If $f:\mathbb R\to\mathbb R$ satisfies

$$f(x+y)=f(x)+f(y)$$

for every $x,y\notin S$, then the same identity holds for every $x,y\in\mathbb R$.

Proof. The equation can fail only on

$$(S\times\mathbb R)\cup(\mathbb R\times S),$$

which is a plane null set. By the main theorem, there is an additive function $h$ such that $f=h$ almost everywhere. Put $k=f-h$, and let

$$T=\{x:k(x)\ne0\},\qquad U=S\cup T.$$

The set $U$ is null. Fix $a\in\mathbb R$. Since $U\cup(a-U)$ is null, choose $a_1$ outside it and put $a_2=a-a_1$. Then $a_1,a_2\notin U$. In particular, $a_1,a_2\notin S$, so the assumed equation gives

$$f(a)=f(a_1)+f(a_2).$$

Because $h$ is additive,

$$k(a)=k(a_1)+k(a_2).$$

Both terms on the right vanish since $a_1,a_2\notin T$. Therefore $k(a)=0$. As $a$ was arbitrary, $f=h$ everywhere, and $f$ is additive everywhere. $\square$

Dependencies. [[analysis/debruijn_1966_almost_additive_functions/main_theorem|Main theorem (Section 2)]].

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