Null master codes and summable slaloms are Tukey equivalent

Enumerate all clopen sets as $(C_i{)}_i$, as in [F1], and put $C_i^k=C_i$ when $\mu (C_i)\le 2^{-k}$ and $C_i^k=\mathrm{\varnothing }$ otherwise. For a null code $f$ define $u(f)(n)=\{f(2n),f(2n+1)\}$. This is a slalom, since ${\sum }_n\mathrm{\mid }u(f)(n)\mathrm{\mid }{2}^{-n}\le {\sum }_n{2}^{1-n}<\mathrm{\infty }$. For $S\in \mathbb{S}$ let $H_S={\mathrm{lim\hspace{0.17em}sup}}_n{\bigcup }_{i\in S(n)}(C_i^{2n}\cup C_i^{2n+1})\mathrm{.}$ Its stage-$n$ measure is at most $\mathrm{\mid }S(n)\mathrm{\mid }(2^{-2n}+2^{-(2n+1)})$, and the sum of these bounds is finite by summability. The elementary tail-union estimate therefore gives $\mu (H_S)=0$. Membership in $H_S$ is Borel in $(S,y)$, since each stage is a finite clopen union.

For a null code $f$, define the closed sets $K_m^{\mathrm{\prime }}(f)={\bigcap }_{n\ge m}(2^{\omega }\setminus C_{f(n)})\mathrm{.}$ They increase to $2^{\omega }\setminus N_f$, which has measure one. Choose the least $m=m(f)$ for which $\mu (K_m^{\mathrm{\prime }}(f))>1\mathrm{/}2$ and set $K_f^{\mathrm{\prime }}=K_{m(f)}^{\mathrm{\prime }}(f)$. This is a Borel choice: each measure is the decreasing limit of measures of finite clopen intersections, and the least-index threshold test is Borel. For the fixed basic clopens $(U_j{)}_j$ let $Z_j=\{f:\mu (K_f^{\mathrm{\prime }}\cap U_j)=0\}$, again Borel by the same finite-stage measure limits, and set $K_f=K_f^{\mathrm{\prime }}\setminus {\bigcup }_{j:f\in Z_j}{U}_j\mathrm{.}$ This is compact, has the same measure as $K_f^{\mathrm{\prime }}$, is disjoint from $N_f$, and has the property that every nonempty $K_f\cap U_j$ has positive measure. The last assertion follows because a zero-measure intersection with $K_f$ would be a zero-measure intersection with $K_f^{\mathrm{\prime }}$ unless $U_j$ was removed, in which case the intersection is empty. Its closed code is Borel in $f$: the finite-stage closed approximants are clopen, and whether the compact intersection meets a basic clopen is the decreasing-limit nonemptiness test from compactness.

Use [F3] to code $\mathbb{S}$ as a Borel subset of a Cantor parameter space; extend the Borel family $H_S$ by empty sections off that subset. Apply [F2] and restrict the resulting selector to obtain a Borel null-code map $v(S)$ with $H_S\subseteq N_{v(S)}$. If $u(f){\subseteq }^{\ast }S$, then for all sufficiently large $n$ the two clopen sets $C_{f(2n)}$ and $C_{f(2n+1)}$ appear in the stage-$n$ union defining $H_S$. Every point of $N_f$ lies in infinitely many even or odd code sets and hence in $H_S$. Thus $N_f\subseteq N_{v(S)}$, proving the first morphism.

For each pair $(n,i)$ with $n\ge 1$, allocate a distinct block of $n$ binary coordinates and let $G_i^n$ be the clopen event that all bits in that block are zero. The blocks are disjoint, so the family of all these events is independent and $\mu (G_i^n)=2^{-n}$. For $S\in \mathbb{S}$ put $L_S={\mathrm{lim\hspace{0.17em}sup}}_{n\ge 1}{\bigcup }_{i\in S(n)}{G}_i^n\mathrm{.}$ The sum of stage measures is bounded by ${\sum }_{n\ge 1}\mathrm{\mid }S(n)\mathrm{\mid }{2}^{-n}<\mathrm{\infty }$, so $L_S$ is null. It is Borel in $(S,y)$. As in step 2.1, [F2] supplies a Borel null-code map $u^{\mathrm{\prime }}(S)$ with $L_S\subseteq N_{u^{\mathrm{\prime }}(S)}$.

For each $f,j,n$ define, when $K_f\cap U_j\mathrm{\ne }\mathrm{\varnothing }$, $T_{f,j}(n)=\{i:K_f\cap U_j\cap G_i^n=\mathrm{\varnothing }\};$ when $K_f\cap U_j=\mathrm{\varnothing }$, put $T_{f,j}(n)=\mathrm{\varnothing }$. The compact-hit tests of step 1.2 make this a Borel family. In the nonempty case put $a=\mu (K_f\cap U_j)>0$. For every finite set of pairs $(n,i)$ with $i\in T_{f,j}(n)$, independence gives $a\le {\prod }_{(n,i)}(1-2^{-n})\mathrm{.}$ Taking finite products increasingly shows both that every $T_{f,j}(n)$ is finite and that ${\sum }_{n\ge 1}\mathrm{\mid }{T}_{f,j}(n)\mathrm{\mid }{2}^{-n}<\mathrm{\infty }$; indeed $-\log (1-t)\ge t$ and the logarithms of all finite products are bounded below by $\log a$. Thus $T_{f,j}$ is a slalom.

Choose Borelly the least increasing thresholds $r_j(f)\ge j$ with ${\sum }_{n\ge r_j(f)}\mathrm{\mid }{T}_{f,j}(n)\mathrm{\mid }{2}^{-n}<2^{-j}$. Such thresholds exist by step 4.1; each test is Borel as a limit of finite sums. Put $v^{\mathrm{\prime }}(f)(n)={\bigcup }_{j:r_j(f)\le n}{T}_{f,j}(n)\mathrm{.}$ Only finitely many $j\le n$ contribute, so the value is finite, and ${\sum }_n\mathrm{\mid }{v}^{\mathrm{\prime }}(f)(n)\mathrm{\mid }{2}^{-n}\le {\sum }_j{2}^{-j}<\mathrm{\infty }$. Also $T_{f,j}{\subseteq }^{\ast }{v}^{\mathrm{\prime }}(f)$ for every $j$.

Assume $N_{u^{\mathrm{\prime }}(S)}\subseteq N_f$. Since $L_S\subseteq N_{u^{\mathrm{\prime }}(S)}$, the compact $K_f$ misses $L_S$. Hence it is covered by the increasing closed sets $K_f\cap {\bigcap }_{n\ge m}(2^{\omega }\setminus {\bigcup }_{i\in S(n)}{G}_i^n)(m\in \omega )\mathrm{.}$ By [F4], one of these closed sets has nonempty relative interior in $K_f$. Choose a basic $U_j$ witnessing that interior. Then $K_f\cap U_j$ is nonempty and, for all $n\ge m$ and $i\in S(n)$, it misses $G_i^n$. Therefore $S(n)\subseteq T_{f,j}(n)$ eventually, and step 5.1 gives $S{\subseteq }^{\ast }{v}^{\mathrm{\prime }}(f)$. This is the second morphism. AC in [F5] supplies DC for the Baire category theorem [F4] at this step and underlies full null-ideal cofinality in [F2]; the displayed Borel maps make no further arbitrary choices. ∎