Lebesgue-point convergence for radial-majorized kernels

Balls are open by the triangle inequality, so F1 establishes measurability before the cube bounds F2 are used. Put $v_n=\mathrm{\mid }B(0,1)\mathrm{\mid }\in (0,\mathrm{\infty })$. F3 gives $\mathrm{\mid }B(0,r)\mathrm{\mid }=v_n{r}^n$ and $\mathrm{\mid }B(0,r)\setminus B(0,r\mathrm{/}2)\mathrm{\mid }=v_n(1-2^{-n})r^n$. On that annulus $\mathrm{\Phi }(\mathrm{\mid }y\mathrm{\mid })\ge \mathrm{\Phi }(r)$; hence $r^n\mathrm{\Phi }(r)\le J_r\mathrm{/}[v_n(1-2^{-n})]\to 0$ as $r\to \mathrm{\infty }$, where $J_r$ is the integral over $\mathrm{\mid }y\mathrm{\mid }\ge r\mathrm{/}2$. Integrability implies these tails tend to zero, by countable additivity on explicit integer annuli. Also disjoint annuli $2^{j-1}\le \mathrm{\mid }y\mathrm{\mid }<2^j$, $j\ge 0$, give ${\sum }_{j\ge 0}{2}^{jn}\mathrm{\Phi }(2^j)\le J\mathrm{/}[v_n(1-2^{-n})]$.

Write $H(y)=\mathrm{\mid }f(x-y)-a\mathrm{\mid }$. Boundedness of $\mathrm{\Phi }$ makes the integral with $f$ absolutely convergent: it is at most ${\epsilon }^{-n}\mathrm{\Phi }(0)\mathrm{\parallel }f{\mathrm{\parallel }}_1$ by reflection and translation of Lebesgue measure. Scaling the defining integral of K gives $\int {\epsilon }^{-n}K(y\mathrm{/}\epsilon )dy=1$. Fact F5 applies to these affine diffeomorphisms and supplies both substitutions. Thus the absolute error is at most $\int {\epsilon }^{-n}\mathrm{\Phi }(\mathrm{\mid }y\mathrm{\mid }\mathrm{/}\epsilon )H(y)dy$.

Given $\eta >0$, fix $\delta >0$ such that $A(r)\le \eta r^n$ for $0<r<\delta$. For $\epsilon <\delta \mathrm{/}2$, the central ball contributes at most $\mathrm{\Phi }(0){\epsilon }^{-n}A(\epsilon )\le \eta \mathrm{\Phi }(0)$. Each dyadic shell $2^j\epsilon \le \mathrm{\mid }y\mathrm{\mid }<2^{j+1}\epsilon$ whose lower radius is below $\delta \mathrm{/}2$ contributes at most ${\epsilon }^{-n}\mathrm{\Phi }(2^j)A(2^{j+1}\epsilon )\le \eta 2^n{2}^{jn}\mathrm{\Phi }(2^j)$. Their sum is bounded by $\eta C$, where $C=\mathrm{\Phi }(0)+2^{n}J\mathrm{/}[v_n(1-2^{-n})]$, independently of $\epsilon$. These regions cover $\mathrm{\mid }y\mathrm{\mid }<\delta \mathrm{/}2$.

On $\mathrm{\mid }y\mathrm{\mid }\ge \delta \mathrm{/}2$, the contribution from $\mathrm{\mid }f(x-y)\mathrm{\mid }$ is at most ${\epsilon }^{-n}\mathrm{\Phi }(\delta \mathrm{/}(2\epsilon ))\mathrm{\parallel }f{\mathrm{\parallel }}_1\to 0$ by step 1.1. The contribution from $\mathrm{\mid }a\mathrm{\mid }$ is at most $\mathrm{\mid }a\mathrm{\mid }{\int }_{\mathrm{\mid }z\mathrm{\mid }\ge \delta \mathrm{/}(2\epsilon )}\mathrm{\Phi }(\mathrm{\mid }z\mathrm{\mid })dz\to 0$ by integrability and scaling. The total error therefore has limit superior at most $\eta C$. Letting $\eta \downarrow 0$ proves convergence. The proof uses the stated countable-choice Euclidean measure interfaces and explicit shells, with no full AC or choice of witnesses at different points.