Chacon levels approximate measurable sets

The algebra of finite unions of intervals in $[0,1)$ with any endpoint conventions generates the trace Borel sigma-algebra by F3. Under AC, F4 replaces $E$ by a Borel set modulo a null set. Given $\epsilon >0$, F2 supplies a finite interval union $V$ with $\mu (E\mathrm{△}V)<\epsilon \mathrm{/}2$. Let $q$ count its finitely many endpoints. The physical partition at stage $r$ has maximum atom length $b_r=\max (w_r,3^{-(r+1)})\to 0$. Outside the at most $2q$ atoms incident to endpoints, every atom is wholly inside or outside $V$. Taking all atoms wholly inside $V$ therefore gives symmetric-difference error at most $2q{b}_r$ (endpoint singletons are null). Removing the reservoir adds at most $3^{-(r+1)}$. Hence a level union $Q_r$ satisfies $\mu (E\mathrm{△}{Q}_r)\le \epsilon \mathrm{/}2+2q{b}_r+3^{-(r+1)}<\epsilon$ for all sufficiently large $r$.

Given $f\in L^2$ and $\epsilon >0$, choose $s={\sum }_{j=1}^m{c}_j{\mathbf{1}}_{E_j}$ with $\mathrm{\parallel }f-s{\mathrm{\parallel }}_2<\epsilon \mathrm{/}2$. If $B=\sum \mathrm{\mid }{c}_j\mathrm{\mid }=0$, use zero. Otherwise approximate each $E_j$ by a level union at one common sufficiently late stage so that every error measure is below $(\epsilon \mathrm{/}(2B){)}^2$, using step 1.1. Then $s_r={\sum }_j{c}_j{\mathbf{1}}_{Q_{r,j}}$ is constant on each level and zero off the tower, and $\mathrm{\parallel }s-s_r{\mathrm{\parallel }}_2\le {\sum }_j\mathrm{\mid }{c}_j\mathrm{\mid }\sqrt{\mu (E_j\mathrm{△}{Q}_{r,j})}<\epsilon \mathrm{/}2$. Thus $\mathrm{\parallel }f-s_r{\mathrm{\parallel }}_2<\epsilon$.

For $\mu (E)>0$ take $0<\eta <\delta \mu (E)\mathrm{/}(1+\delta )$ and choose, at any sufficiently late stage, a level union $Q$ with $\mu (E\mathrm{△}Q)<\eta$. Then $\mu (Q)>\mu (E)-\eta >0$. If every level of $Q$ had $E$-proportion at most $1-\delta$, summing over its disjoint levels would give $\mu (Q\setminus E)\ge \delta \mu (Q)>\delta (\mu (E)-\eta )>\eta$, contradicting the error bound. At least one level has the required strict density. The case $\delta \ge 1$ is also covered by this contradiction, and only finitely many levels are compared.