Corollary (Section 6)

---

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

Statement. In the setting of Theorem 3, suppose $\mu(G)=\infty$. If

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

outside a subset of $G\times G$ of finite outer product measure, then $f$ is almost everywhere equal to a homomorphism. Consequently the displayed equation itself holds almost everywhere.

For $G=H=\mathbb R$ with Lebesgue measure, this weakens the hypothesis of Erdős Problem 1126 from a plane null exceptional set to one of arbitrary finite outer plane measure, and so strengthens the result.

Proof. Choose a finite positive $\beta>0$ that bounds the outer product measure of the exceptional set. This also covers a null exceptional set. When $\mu(G)=\infty$, every $\alpha>0$ satisfies the three inequalities (8) in Theorem 3. Therefore, for each $\alpha>0$, there is a homomorphism $h_\alpha$ such that

$$\mu^*\{x:f(x)\ne h_\alpha(x)\}\leq\alpha.$$

The homomorphism does not depend on $\alpha$. Indeed, if $h_\alpha$ and $h_\gamma$ are two such homomorphisms, let $E$ be the union of their two disagreement sets with $f$. The set $E$ has finite outer measure. For any $t\in G$, the set $E\cup(t-E)$ still has finite outer measure and hence cannot be all of the infinite-measure group $G$. Choose $y\notin E\cup(t-E)$. The two homomorphisms agree both at $y$ and at $t-y$, so additivity shows that they agree at $t$.

Fix the common homomorphism $h$. Its disagreement set with $f$ has outer measure at most $\alpha$ for every $\alpha>0$, and therefore has outer measure zero. This proves $f=h$ almost everywhere.

For the Lebesgue case $G=\mathbb R$, let $E=\{x:f(x)\ne h(x)\}$. The original equation can fail only on

$$(E\times\mathbb R)\cup(\mathbb R\times E) \cup\{(x,y):x+y\in E\}.$$

The first two sets are plane null by Fubini's theorem, and the third is plane null under the measure-preserving shear $(x,y)\mapsto(x,x+y)$. Thus the equation holds almost everywhere in the case relevant to Problem 1126. $\square$

Proof coverage. The correction $f=h$ almost everywhere is a complete deduction conditional on Theorem 3. The last three-null-set argument above is complete for Lebesgue measure on $\mathbb R$. At the source's general measurable-group breadth, the corresponding product-null assertion depends on the same unexpanded product-measure conventions recorded as a gap for Theorem 3; no source error is asserted.

Dependencies. Theorem 3.

Bears on. #1126