Erdős Problem #908
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Erdős problems openaigpt 6 astra ultra 100% cataloguedSolvedSolvedStatement
The current problem page assumes that every difference $\Delta_t f$ is measurable, but asks for $f=g+H+r$ with $g$ continuous, $H$ additive, and $\Delta_t r=0$ almost everywhere for each fixed $t$. That literal continuous-$g$ statement is false. Laczkovich's original Theorem 3 instead has $g$ Lebesgue measurable. We first give the counterexample to the literal wording, then reconstruct the corrected theorem relative to two explicitly stated preparatory results from the same paper.
Restatement from Math-5985-A1 r1, with the source manuscript’s qualifications.
Original problem on erdosproblems.com
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GPT 6 Astra Ultra attempt on Math-5985
Erdős Problem 908: the measurable conclusion and a wording correction
Literal statement and the primary theorem.
The current problem page assumes that every difference $\Delta_t f$ is measurable, but asks for $f=g+H+r$ with $g$ continuous, $H$ additive, and $\Delta_t r=0$ almost everywhere for each fixed $t$. That literal continuous-$g$ statement is false. Laczkovich's original Theorem 3 instead has $g$ Lebesgue measurable. We first give the counterexample to the literal wording, then reconstruct the corrected theorem relative to two explicitly stated preparatory results from the same paper.A counterexample to continuity of $g$.
Take $f(x)=\mathbf1_{[0,\infty)}(x)$. It is measurable, so all its differences are measurable. Its difference with shift one equals one on $(-1,0)$ and zero on $(-\infty,-1)\cup(0,\infty)$. If the displayed continuous-$g$ decomposition existed, then for almost every $x$ \[ \Delta_1 f(x)=g(x+1)-g(x)+H(1). \] The right side is continuous. A continuous function equal almost everywhere to a constant on an open interval equals that constant everywhere there: otherwise a neighbourhood of positive measure would contradict the almost-everywhere equality. It would therefore equal zero just to the left of $-1$ and one just to its right, an impossibility. This argument does not assume that $r$ is measurable and uses only the null set associated with shift one.Two explicit inputs for the corrected theorem.
On measurable 1-periodic functions modulo equality almost everywhere, use the translation-invariant metric \[ \|u\|_0=\int_0^1\min(1,|u(x)|)\,dx,\qquad s(u)=\inf_{a\in\mathbb R}\|u-a\|_0. \] Convergence for this metric is convergence in measure. We retain the following two results of Laczkovich as external inputs, stated in their equivalent convergence-in-measure form.- Theorem 2: if $u_j$ are measurable 1-periodic functions and $\Delta_tu_j\to0$ in measure for every fixed real $t$, then $s(u_j)\to0$.
- Theorem 1: if $c:\mathbb R\to\mathbb R$, $c(0)=0$, and \[ c(h+k)-c(h)-c(k)\longrightarrow0\quad((h,k)\to(0,0)), \] then $c=H+u$, where $H$ is additive and $u(h)\to0$ as $h\to0$, with $u(0)=0$.
Reduction to a periodic function.
All real shifts have measurable differences, since $\Delta_{-t}f(x)=-\Delta_tf(x-t)$. Define $F$ by $F(x)=f(x)$ on $[0,1)$ and extend it with period one. On each interval $[j,j+1)$, $f(x)-F(x)=f(x)-f(x-j)$ is measurable. Thus $f-F$ is measurable, and every difference of $F$ is measurable. It suffices to decompose $F$. Put $D_h(x)=F(x+h)-F(x)$.For $h_j\to0$ and fixed $t$, \[ \Delta_tD_{h_j}(x)=\Delta_{h_j}D_t(x)\longrightarrow0 \quad\hbox{in measure}. \] The first external input gives $s(D_h)\to0$ as $h\to0$. Choose a minimizing constant $c(h)$, so that $\|D_h-c(h)\|_0=s(D_h)$, choosing periodically and with $c(0)=0$. Such a minimum exists: the metric is continuous in the constant and tends to one as its absolute value tends to infinity; if the infimum is one, every constant minimizes.
The identity \[ D_{h+k}(x)=D_h(x+k)+D_k(x) \] and the triangle inequality imply \[ \min(1,|c(h+k)-c(h)-c(k)|) \leq s(D_{h+k})+s(D_h)+s(D_k)\longrightarrow0. \] Apply the second external input to obtain $c=H+u$. Replacing $H(x)$ by $H(x)-xH(1)$ and $u(x)$ by $u(x)+xH(1)$, we may suppose $H(1)=0$.
A jointly measurable representative.
Set \[ K(x,h)=F(x+h)-F(x)-H(h). \] For fixed $h$, this is measurable and 1-periodic in $x$. Moreover \[ \|K(\,\cdot\,,h)\|_0 \leq s(D_h)+\min(1,|u(h)|)\longrightarrow0 \quad(h\to0). \] Its cocycle identity gives \[ \|K(\,\cdot\,,h)-K(\,\cdot\,,h')\|_0 =\|K(\,\cdot\,,h-h')\|_0. \] Choose positive meshes $\eta_j\to0$ so small that the last norm is below $2^{-3j}$ whenever $|h-h'|\leq\eta_j$, and define \[ G_j(x,h)=K(x,\eta_j\lfloor h/\eta_j\rfloor). \] Each $G_j$ is jointly Lebesgue measurable, and its section as a function of $h$ is Borel for every $x$. For fixed $h$, \[ \lambda\{x\in[0,1]:|G_j(x,h)-K(x,h)|\geq2^{-j}\} \leq2^{-2j}. \] Borel--Cantelli shows that $G_j(x,h)\to K(x,h)$ for almost every $x$, for each fixed $h$. Define $G(x,h)$ as the finite limit when it exists, and zero otherwise. Then $G$ is jointly measurable, each $G(x,\cdot)$ is Borel, and \[ R(x,h):=K(x,h)-G(x,h)=0 \quad\hbox{for a.e. }x,\quad\hbox{for every fixed }h. \tag{1} \] Periodicity in $x$ extends this assertion from $[0,1]$ to the whole line.Extracting the almost-invariant remainder.
Although joint measurability of $R$ is not presumed, its defect is jointly measurable: \[ \begin{split} L(x,y,z) &:=R(x,y+z)-R(x+y,z)-R(x,y)\\ &=-G(x,y+z)+G(x+y,z)+G(x,y). \end{split} \] For fixed $y,z$, $1$ makes $L(x,y,z)=0$ for a.e. $x$. Fubini therefore supplies $x_0$ such that $L(x_0,y,z)=0$ for almost every pair $(y,z)$. For almost every $z$, this and $1$ give \[ R(x_0,y+z)-R(x_0,y)=0\quad\hbox{for a.e. }y. \tag{2} \] Let $Z$ be the full-measure set of such $z$. For any real $t$, choose $z_1,z_2\in Z$ with $t=z_1+z_2$; this is possible because $Z\cap(t-Z)$ has full measure. Adding the two translated instances of $2$ proves the same equality for shift $t$. The exceptional null set may depend on $t$, exactly as required in the question.Now put \[ r(x)=R(x_0,x-x_0),\qquad g_0(x)=G(x_0,x-x_0)+F(x_0)-H(x_0). \] By construction $g_0$ is measurable, $F=g_0+H+r$, and $\Delta_t r=0$ a.e. for every $t$. Finally $g=g_0+(f-F)$ is measurable and $f=g+H+r$, proving the corrected theorem.