GPT 6 Astra Ultra attempt on Math-6180

Erdős Problem 1126: correcting an almost additive function

Statement and scope.

If $f:\mathbb R\to\mathbb R$ satisfies $f(x+y)=f(x)+f(y)$ for almost every pair $(x,y)$, is it almost everywhere equal to an additive function? Yes. Here is a reconstruction of the known de Bruijn--Jurkat conclusion which does not assume that $f$ is measurable. The source is https://www.erdosproblems.com/1126.

Proof.

The hypothesis means that the exceptional pairs lie in a measurable planar null set $N$. Fubini's theorem, applied to its indicator, gives a conull set $E\subset\mathbb R$ such that \[ f(y+t)-f(y)=f(t)\quad\hbox{for almost every }y \qquad(t\in E). \] For any $t\in\mathbb R$, choose $a\in E\cap(t-E)$, and put $b=t-a\in E$. The preceding identities for $a,b$, together with translation invariance of null sets, imply \[ f(y+t)-f(y) =[f(y+b+a)-f(y+b)]+[f(y+b)-f(y)] =f(a)+f(b) \] for almost every $y$. Thus the translation difference by $t$ is almost everywhere a constant. Define $g(t)$ to be that constant. It is unique: two conull sets of valid $y$'s intersect, so different choices of $a,b$ give the same value.

For arbitrary $s,t$, intersect the three conull sets on which the identities for the translations $s,t,s+t$ hold (translating the first set as appropriate). At any point of this intersection, \[ g(s+t)=f(y+s+t)-f(y) =[f(y+t+s)-f(y+t)]+[f(y+t)-f(y)] =g(s)+g(t). \] Hence $g$ is additive everywhere. For $t\in E$, its defining constant is $f(t)$, so $f=g$ almost everywhere.

The additive representative is also unique. If additive $g_1,g_2$ agree almost everywhere, then for any $t$ choose $y$ such that they agree at both $y$ and $y+t$; subtracting gives $g_1(t)=g_2(t)$.

Attribution and dependencies.

The database attributes the general theorem to de Bruijn $1966$ and Jurkat $1965$. The only external analytic input used above is Fubini's null-section theorem and translation invariance of Lebesgue null sets. There is no appeal to integration or measurability of $f$, and no assertion that the resulting additive function must be linear.

Completion estimate.

The full stated existence and uniqueness assertion is proved. COMPLETION: 100%