Theorem 3

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Source. Section 6, printed p. 62 (PDF p. 4).

Statement. Let $G$ be a measurable abelian group with a measure $\mu$ satisfying

$$0<\mu(G)\leq\infty$$

and invariant under $x\mapsto x+a$ and $x\mapsto-x$. In the group-valued setting of Section 4, let $H$ be an additive abelian group and let $f:G\to H$. Let $\alpha,\beta$ be finite positive numbers such that

$$2\alpha<\mu(G),\qquad 3\beta<\alpha\mu(G),\qquad 2\beta< \left(\mu(G)-\frac{2\beta}{\alpha}\right) \left(\mu(G)-\frac{4\beta}{\alpha}\right). \tag{8}$$

Suppose

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

for every pair $(x,y)$ outside an exceptional set $N\subseteq G\times G$ with product outer measure at most $\beta$. Then there is a homomorphism $h:G\to H$ such that $f(x)=h(x)$ outside a subset of $G$ of outer measure at most $\alpha$.

Proof sketch. Let $N$ be the exceptional subset of $G\times G$. The product-measure estimate gives a set $M\subseteq G$ of outer measure at most $\alpha$ such that

$$\mu^*(N_x)\leq\frac{\beta}{\alpha}$$

for every $x\notin M$. The first inequality in (8) ensures that $M\cup(x-M)\ne G$. Choosing $x_1$ outside this union, as in Section 2, defines $h(x)$ so that

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

outside a set of $y$'s of outer measure at most $2\beta/\alpha$.

For $x\notin M$, the original equation and the displayed identity have a common valid $y$: their combined exceptional outer measure is at most $3\beta/\alpha<\mu(G)$. Consequently $h(x)=f(x)$, so the disagreement set has outer measure at most $\alpha$.

To prove additivity, the five equations from Section 2 must again hold simultaneously. The first restricts $w$ by a set of outer measure at most $2\beta/\alpha$. For every remaining $w$, the next two equations restrict $z$ by outer measure at most $4\beta/\alpha$. The allowed set of $z$'s depends on $w$, since one condition has the form $w+z\notin K_{a+b}$; the candidate region is therefore described fiberwise, not as a rectangle.

The paper uses these bounds to give the candidate pairs the lower product bound

$$\left(\mu(G)-\frac{2\beta}{\alpha}\right) \left(\mu(G)-\frac{4\beta}{\alpha}\right).$$

The original exceptional set and its translate together have product outer measure at most $2\beta$. The last inequality in (8) makes the displayed lower bound larger than $2\beta$, so the paper concludes that one pair $(w,z)$ satisfies all five equations. Their cancellation gives $h(a+b)=h(a)+h(b)$.

Proof coverage. The paper gives the quantitative bounds above and refers to the equations and cancellation in Section 2. It does not specify the measurability and product-measure conventions needed to pass from the varying fiberwise outer-measure bounds to the displayed lower product bound in the stated general measurable group. This page preserves the source's argument but does not claim a full arbitrary-group measure-theoretic reconstruction. That reconstruction remains a gap; no source error is asserted.

Dependencies. The five-equation argument in [[analysis/debruijn_1966_almost_additive_functions/main_theorem|the main theorem]].

Bears on. #1126