Kolmogorov atomic kernel maxima

At step $j$, the rational linear span of $1,t_1,\dots ,t_{j-1}$ is countable: enumerate rational tuples by listing integer numerators and positive denominators of bounded absolute size, then increase the bound; evaluate each tuple. F3 makes this span null. The open cell $((j-1)\mathrm{/}n,j\mathrm{/}n)$ has measure $1\mathrm{/}n>0$, so contains a point $t_j$ outside the span. Finite induction chooses these points and proves their independence. Let $V$ be their full rational linear span with one; the same enumeration makes $V$ countable and null. AC supplies the countable-choice measure hypotheses in F3; selecting this fixed finite tuple adds no arbitrary-index choice.

Fix $x\in [0,1)\setminus V$ and write $y_j=x-t_j$. If $q_0+{\sum }_j{q}_j{y}_j=0$ for rational coefficients, put $q={\sum }_j{q}_j$. If $q=0$, independence of $1,t_1,\dots ,t_n$ forces all coefficients zero. If $q\mathrm{\ne }0$, the equation puts $x=({\sum }_j{q}_j{t}_j-q_0)\mathrm{/}q$ in $V$, impossible. Thus $1,y_1,\dots ,y_n$ are independent, and none of the $y_j$ is an integer.

Set $z_j=i{e}^{-\pi i{y}_j}\mathrm{sgn}(\sin (\pi y_j))$, which has modulus one. Apply F1 with error $1\mathrm{/}2$ to obtain $L\ge 1$. Multiplying the approximation by $e^{\pi i{y}_j}\mathrm{sgn}(\sin (\pi y_j))$ shows $\mathrm{\mid }{e}^{(2L+1)\pi i{y}_j}\mathrm{sgn}(\sin (\pi y_j))-i\mathrm{\mid }<1\mathrm{/}2$. Its imaginary part exceeds $1\mathrm{/}2$, hence F2 gives $D_L(y_j)>1\mathrm{/}(2\mathrm{\mid }\sin (\pi y_j)\mathrm{\mid })\ge 1\mathrm{/}(2\pi \mathrm{\mid }{y}_j\mathrm{\mid })$. All terms have the same positive sign; there is no cancellation in their average.

Let $k\in \{1,\dots ,n\}$ be the cell index with $(k-1)\mathrm{/}n\le x<k\mathrm{/}n$. Cell placement implies $\mathrm{\mid }x-t_j\mathrm{\mid }\le (\mathrm{\mid }j-k\mathrm{\mid }+1)\mathrm{/}n$. At least one side from $k$ to an endpoint contains $m+1\ge n\mathrm{/}2$ indices with distances $\mathrm{\ell }=0,\dots ,m$. Therefore $n^{-1}{\sum }_j\mathrm{\mid }x-t_j{\mathrm{\mid }}^{-1}\ge {\sum }_{\mathrm{\ell }=0}^m(\mathrm{\ell }+1{)}^{-1}\ge \log (m+2)\ge \log (n\mathrm{/}2)\ge (\log n)\mathrm{/}2$. The harmonic bound follows by integrating $1\mathrm{/}t$ over each $[\mathrm{\ell }+1,\mathrm{\ell }+2]$, where it is at most $1\mathrm{/}(\mathrm{\ell }+1)$. Step 3.1 now gives $B_L(x)>(\log n)\mathrm{/}(4\pi )$. This holds off the single countable null set $V$, proving the assertion with the stated uniform constant. Grid endpoints are covered by the half-open choice of k; singular points $t_j$ were excluded in step 2.1.