Jurkat 1965 cauchy functional equation

../

theorem_i: An almost-everywhere solution of Cauchy's equation agrees almost everywhere with a unique function that is additive on the whole real line.

theorem_ii: An everywhere additive real function with f(1/x) = f(x)/x^2 for every nonzero x is linear, f(x) = x f(1) for all real x.

---

Jurkat, Wolfgang B., "On Cauchy's Functional Equation." Proceedings of the American Mathematical Society 16, no. 4 (1965), 683--686. DOI: 10.1090/S0002-9939-1965-0179496-8.

Theorem I independently solves Erdős's Problem P 310. It allows the original real-valued function to be undefined on a linear null set. If Cauchy's equation holds for almost every pair in the sense of two-dimensional Lebesgue measure, the theorem constructs a unique everywhere-defined additive function that agrees with the original function almost everywhere.

Jurkat first applies Fubini's theorem to choose a conull set $M$ on which the equation can be composed safely. Three null-set avoidances show that the equation holds whenever $x,y,x+y\in M$. He then proves that $f(x)+f(y)$ depends only on $x+y$ for $x,y\in M$, and proves a three-to-two summand reduction inside $M$. Since every real number is a sum of two elements of $M$, these facts define a function $F$ on all of $\mathbb R$ and prove its additivity. A final conull decomposition proves uniqueness.

This is materially different from de Bruijn's proof in [[analysis/debruijn_1966_almost_additive_functions/main_theorem|Section 2 of de Bruijn 1966]]. Both arguments begin with Fubini's theorem and finite unions of translated null sets. Jurkat extends $f$ through consistent representations in $M+M$; de Bruijn instead defines $h(x)$ as the almost-everywhere constant value of $f(x+y)-f(y)$, then uses a five-exception argument in the $(w,z)$-plane. Jurkat's added-in-proof note says that de Bruijn sent him the independent manuscript in September 1964.

Theorem II answers a question of I. Halperin: an everywhere additive real function with $f(1/x)=f(x)/x^2$ for every $x\ne0$ satisfies $f(x)=xf(1)$ for every real $x$, with no regularity assumed. It does not bear on Problem 1126.

Read status. Claims checked: the statements of Theorems I and II were read clause by clause against the print, and their proofs were read through but not verified by a second reader.

Copy read. The copy read for this card is the AMS article PDF from https://www.ams.org/journals/proc/1965-016-04/S0002-9939-1965-0179496-8/S0002-9939-1965-0179496-8.pdf. The 1965 pages print no copyright line; the publisher's article page rendered only the site shell, with no copyright line (https://pubs.ams.org/journals/proc/1965-016-04/S0002-9939-1965-0179496-8, read 2026-10-02), the Crossref record names no license, and the publisher's copyright policy page states that authors transfer copyright to the American Mathematical Society and that Creative Commons licenses apply only to its open-access series (https://www.ams.org/publications/authors/ctp, read 2026-10-02), every other right reserved.

Bears on. #1126: Theorem I proves the problem's statement as the problem page formulates it, with almost all pairs taken in two-dimensional and the almost-everywhere agreement in one-dimensional Lebesgue measure; it allows $f$ to be undefined on a null set and adds that the additive function is unique. Theorem II bears on no Erdős problem.

Results.

• Theorem I: existence and uniqueness of the everywhere additive correction.
• Theorem II: an additive function with $f(1/x)=f(x)/x^2$ is $x f(1)$.

No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.