Solovay densities and localized small null joins

First reconstruct the integral foundation used by [F3]–[F7]. Augment every finite disjoint display $s={\sum }_r{a}_r{\mathbf{1}}_{E_r}$ of a nonnegative simple function by $X\setminus {\bigcup }_r{E}_r$ with coefficient $0$. Intersections of two augmented displays partition $X$, and equality of the functions makes their coefficients equal on every nonempty cell. Finite additivity and $0\cdot (+\mathrm{\infty })=0$ therefore prove representation independence. Common augmented refinements give simple monotonicity and additivity termwise; homogeneity is direct for scalar $0$ and termwise for a positive scalar. Taking suprema over simple minorants gives nonnegative monotonicity. If $0\le f_j\uparrow f$, put $L={\sup }_j\int f_j$. For every simple $s\le f$ and $0<c<1$, the sets $A_j=\{f_j\ge cs\}$ increase to $X$, including on the zero level of $s$, and continuity from below for the finite-sum measure $A\mapsto {\int }_{A}s$ gives $c\int s={\lim }_{j}c{\int }_{A_j}s\le L$. Let $c\uparrow 1$ and take the supremum over $s$ to obtain MCT. Applying MCT to sums of increasing simple approximants gives nonnegative additivity; positive/negative and real/imaginary decompositions then give finite $L^1$ linearity. Finally, if $g\ge 0$ and $\int g=0$, then $(1\mathrm{/}n)\mu \{g\ge 1\mathrm{/}n\}\le \int g$ for every $n$, so $g=0$ almost everywhere; the converse follows because every simple minorant is supported, apart from its zero cell, on a null set. These arguments supply the exact affected parts of [F4]–[F7], and with them substituted at its base the unaffected RN construction and uniqueness argument in [F3] applies.

Every function $v:I\to [0,1]$ has exactly one $U$-large fibre. If none were large, the complements of all its fibres would belong to $U$. The range has cardinality below $\kappa$, so indexing it by an ordinal below $\kappa$ using F8 and applying F2 would put their empty intersection in the proper ultrafilter. Two disjoint fibres cannot both belong to a proper filter. Apply this to $v(i)=m(c\wedge a_i)$ for each $(a,c)$ to define the unique number ${\nu }_a(c)$. Uniqueness and Replacement give a set function of $(a,c)$, with $0\le {\nu }_a(c)\le m(c)$ and ${\nu }_a(0)=0$.

Fix $a$ and pairwise disjoint $c_n\in \mathbb{B}$, and put $c={\bigvee }_n{c}_n$. Intersect the $U$-large fibres defining ${\nu }_a(c)$ and all ${\nu }_a(c_n)$. This countable intersection is in $U$ by F2 and is nonempty. At any $i$ in it, meet-distributivity and countable additivity from F1 give $m(c\wedge a_i)={\sum }_{n}m(c_n\wedge a_i)$, hence ${\nu }_a(c)={\sum }_n{\nu }_a(c_n)$. Thus $E\mapsto {\nu }_a([E])$ is a finite positive measure on $\mathrm{\Sigma }$, dominated by $\mu$. It is absolutely continuous, and its total variation equals itself: each measurable finite partition sums to the measure of its union because all values are nonnegative. F3 therefore applies with the constant finite exhaustion $X$.

Let $h$ be the integrable real RN density from step 2.1. For each positive integer $n$, put $E_n=\{h\le -1\mathrm{/}n\}$ and $T_n=\{h\ge 1+1\mathrm{/}n\}$. Linearity and monotonicity give ${\nu }_a([E_n])={\int }_{E_n}hd\mu \le -\mu (E_n)\mathrm{/}n$ and ${\nu }_a([T_n])\ge (1+1\mathrm{/}n)\mu (T_n)$. Since $0\le {\nu }_a([E_n])$ and ${\nu }_a([T_n])\le \mu (T_n)$, both sets are null. Their countable union contains the set where $h\mathrm{\notin }[0,1]$. Replacing $h$ by $\min (1,\max (0,h))$ gives a measurable $[0,1]$-valued density; integration is unchanged on the null exceptional set. F3 gives almost-everywhere uniqueness. The set of all measurable functions $X\to [0,1]$ is a subset of $[0,1{]}^X$; each equivalence class of densities is thus a nonempty set, uniquely specified by $a$. Replacement gives the class assignment as a set function, and F8 permits simultaneous representatives if desired. Integrals over equivalent measurable representatives of $c$ agree, because these functions are bounded and the symmetric difference is null.

Suppose the hypothesis of locality holds on $J\in U$, and let $d\le c$. For $i\in J$, $d\wedge a_i=d\wedge b_i$. Intersecting $J$ with the two defining large fibres in step 1.2 proves ${\nu }_a(d)={\nu }_b(d)$. Choose a measurable representative $C$ of $c$. Consequently $1_C{h}_a$ and $1_C{h}_b$ have identical integrals over every $E\in \mathrm{\Sigma }$, since those integrals equal ${\nu }_a([E]\wedge c)$ and ${\nu }_b([E]\wedge c)$. Both are densities of the same finite measure, so F3 gives their equality almost everywhere, which is exactly equality on $c$.

For the indicator family of $Z$, the values $m(c\wedge a_i)$ are $m(c)$ on $Z$ and zero off $Z$. The ultrafilter decides $Z$, so step 1.2 gives ${\nu }_a(c)=m(c)$ or zero accordingly, including $c=0$. Constant densities $1$ and $0$ represent these measures, so uniqueness gives the assertion. For complements, $m(c\wedge \mathrm{\neg }{a}_i)=m(c)-m(c\wedge a_i)$ for every $i$, and the large-fibre equation yields ${\nu }_{\mathrm{\neg }a}(c)=m(c)-{\nu }_a(c)$. F4 shows that $1-h_a$ represents this measure, so uniqueness gives the complement identity.

For the sequence in clause 3 and any $d\le c$, the family $(d\wedge a_i^n{)}_n$ is disjoint for every $i$. Meet-distributivity and countable additivity give $m(d\wedge a_i)={\sum }_{n}m(d\wedge a_i^n)$. Intersect the countably many defining large fibres, including that for $a$, to obtain ${\nu }_a(d)={\sum }_n{\nu }_{a^n}(d)$. Choose nonnegative representatives of all these densities by F8. For a measurable representative $C$ of $c$, F4 and F6 applied to $1_C{\sum }_{n<N}{h}_{a^n}$ show that its increasing pointwise limit has integral over $E$ equal to ${\nu }_a([E]\wedge c)$. Its total integral is at most $1$, so the set where the limit is infinite is null: on that set every constant bound has integral at most $1$, forcing its measure to be zero. Replace the limit there by zero to obtain a real integrable density of the same finite measure as $1_C{h}_a$. F3 gives equality almost everywhere, proving clause 3 and the asserted finiteness on $c$.

For clause 4, the hypotheses and F7 give ${\nu }_{a^{\xi }}(c)=0$ for every $\xi <\beta$. By step 1.2 each set $J_{\xi }=\{i:m(c\wedge a_i^{\xi })=0\}$ belongs to $U$. Their intersection $J$ is in $U$ by F2. For $i\in J$, strict positivity in F1 makes every $c\wedge a_i^{\xi }$ the zero Boolean element. Meet-distributivity now gives $c\wedge a_i={\bigvee }_{\xi <\beta }(c\wedge a_i^{\xi })=0$. Thus ${\nu }_a(c)=0$ by the unique large-fibre rule, and F7 gives $h_a=0$ almost everywhere on $c$. For $\beta =0$, $J=I$, every $a_i=0$ and step 4.2 gives the zero density directly. No union of fewer than $\kappa$ measurable exceptional sets has been formed.

The zero set of each density is measurable and its class is independent of its representative, by almost-everywhere uniqueness. Put $c={\bigwedge }_{\xi <\beta }z(a^{\xi })$ using F1. For each $\xi$, the inequality $c\le z(a^{\xi })$ means precisely that $h_{a^{\xi }}$ vanishes almost everywhere on a representative of $c$. Step 5.1 then gives $c\le z(a)$. This includes the empty meet $c=1$, the singleton family and the zero condition. All constructions and identities concern sets and functions in the original universe; no generic interpretation has entered the argument.