GPT 6 Astra Ultra attempt on Math-5984

Erdős Problem 907: continuous differences

Statement and approach.

If $f:\mathbb R\to\mathbb R$ has continuous differences $\Delta_t f(x)=f(x+t)-f(x)$ for every $t>0$, then $f=g+H$, where $g$ is continuous and $H$ is additive. We prove the reduction to a precise measurable-cocycle theorem and then supply the regularity argument. The cocycle theorem is an explicit external dependency.

From separate continuity to a measurable defect.

Negative shifts cause no difficulty, since $\Delta_{-t}f(x)=-\Delta_t f(x-t)$. Define \[ D(x,y)=f(x+y)-f(x)-f(y). \] For fixed $y$, this is continuous in $x$; symmetry gives continuity in $y$ for fixed $x$. To check joint measurability without assuming $f$ itself measurable, put \[ D_j(x,y)=D(2^{-j}\lfloor2^jx\rfloor,y). \] Each $D_j$ is Borel measurable: it is defined on countably many Borel vertical strips by continuous functions of $y$. Separate continuity gives $D_j(x,y)\to D(x,y)$ at every point. Hence $D$ is jointly Borel, in particular Lebesgue measurable.

The precise external theorem.

Theorem 5 of M. Laczkovich, Functions with measurable differences, Acta Math. Acad. Sci. Hungar. 35 $1980$, 217--235, states that if $f(x+y)-f(x)-f(y)$ is jointly Lebesgue measurable, then $f=u+H$, with $u$ Lebesgue measurable and $H$ additive. We apply exactly this theorem. It makes no preliminary measurability assumption on $f$. Its proof, which builds on the weak difference-property theorem, is not assumed to have been reproduced here. Every difference of $u=f-H$ is continuous, because $\Delta_tu=\Delta_tf-H(t)$.

A measurable function with continuous differences is continuous.

We prove the needed regularity assertion. Let $x_j\to0$ be any sequence, and set \[ c_j=u(x_j)-u(0),\qquad E_j(t)=u(t+x_j)-u(t)-c_j. \] For fixed $t$, \[ E_j(t)=\Delta_tu(x_j)-\Delta_tu(0)\longrightarrow0. \] The measurable functions $E_j$ therefore converge to zero in measure on $[0,1]$. Also translation continuity in measure gives \[ u(t+x_j)-u(t)\longrightarrow0 \quad\text{in measure on }[0,1]. \tag{1} \] It follows that the constant functions $c_j$ tend to zero in measure, and hence $c_j\to0$ as real numbers. Thus $u$ is continuous at zero. For every fixed $a$, \[ u(a+x_j)-u(a) =u(x_j)-u(0)+\Delta_au(x_j)-\Delta_au(0)\longrightarrow0, \] so $u$ is continuous everywhere. Taking $g=u$ proves the desired decomposition.

Here $1$ requires only that $u$ be finite and Lebesgue measurable. One direct proof is to approximate its restriction to $[-1,2]$, outside a set of arbitrarily small measure, by a continuous function, using Lusin's theorem and extension from a compact set. For small $x_j$, both $t$ and $t+x_j$ lie in that interval. The exceptional sets have arbitrarily small total measure, and the continuous approximation has uniformly small translation differences.

The ambiguity is exactly linear.

If $f=g_1+H_1=g_2+H_2$ are two such decompositions, then $H_1-H_2=g_2-g_1$ is both continuous and additive. Its value on rational $q$ is $q(H_1-H_2)(1)$; density of the rationals gives the same formula for all reals. Thus the only ambiguity is moving a linear function between the two summands.

Attribution and assessment.

The original resolution is de Bruijn's 1951 theorem: https://pure.tue.nl/ws/files/4316577/597498.pdf. The later measurable-cocycle input was checked directly on page 229 of Laczkovich's original paper, archived in https://real-j.mtak.hu/7443/1/MTA_ActaMathHung_35.pdf. See also https://www.erdosproblems.com/907, checked 24 September 2026. This is a complete deduction relative to that explicit input and standard measure theory, not a claim of a new proof without external dependencies or of Lean verification.

Completion estimate.

COMPLETION: 100%