All sets of reals in Solovay L(R) have LM, BP, and PSP

If $A\subseteq \mathbb{R}$ lies in $K$, choose $(\alpha ,r)$ with $A=F(\alpha ,r)$. Thus membership in $A$ is expressed by the canonical hierarchy definition from the single real $r$ and ordinal $\alpha$; no unlisted earlier-stage parameters remain. Localize $r$ by F4. Since $L(\mathbb{R})$ is canonically definable from the class of all reals, the remaining homogeneous forcing fixes the membership formula. F2 gives Borel $B$ with $A\mathrm{△}B$ contained in a coded null set, and likewise an open representative modulo a coded meagre set. All witness codes are reals and hence lie in $K$ by F1. Absoluteness and internal DC therefore give LM and BP in $K$.

Use F8's half-open binary coding $b:[0,1)\to 2^{\omega }$. Let $D$ be the Borel conull set of $x$ for which none of the three residue-class subsequences $k\mapsto b(x)(3k+j)$ is eventually $1$. Splitting those subsequences and decoding them gives a Borel bijection $T:D\to [0,1{)}^3$; its inverse interleaves the three canonical codes. A length-$3k$ cylinder has measure $2^{-3k}$ and maps to a product of three length-$k$ dyadic intervals, also of measure $2^{-3k}$. The monotone-class argument from these generating cylinders, followed by the product-completion theorem in F8, shows that $T$ and $T^{-1}$ preserve Borel sets and send Borel null sets to Borel null sets. This conclusion is derived here, not attributed to the one-way statement of the dyadic lemma.

Let $N=V[G_{\xi }]$ be the bounded intermediate stage containing $r$. By F4, $N\cap \mathbb{R}$ has in the final extension an enumeration coded by a real, so that enumeration belongs to $K$ by F1. If $A$ is uncountable in $K$, some $z\in A\setminus N$ therefore exists. Localize $z$ to $V[G_{\eta }]=N[H]$ for some $\eta >\xi$ and choose in $N$ a name $\dot{\tau }$ for $z$ over the interval collapse $Q$.

In $N$ let $D=\{q\in Q:(\mathrm{\exists }y\in {\mathbb{R}}^N)q⊩\dot{\tau }=\check{y}\}$. The actual generic $H$ misses $D$. Since $D\cup \{q:q\perp D\}$ is a dense set of $N$, some $p_0\in H$ is incompatible with $D$, so no extension of $p_0$ forces a ground-real value. The truth lemma and homogeneous tail forcing give $p_1\in H$ forcing the canonical membership formula from step 1.1. Take $p\in H$ below both. Apply F3 below $p$: every branch interpretation satisfies that membership formula, and the resulting injective continuous image is perfect. Its tree and image codes are reals and hence belong to $K$. Thus $A$ has a perfect subset. This uses the forcing predicate only on set parameters in $N$, never the external formula “$\dot{\tau }\mathrm{\notin }N$.” If no such $z$ exists, the displayed enumeration instead proves $A$ countable.

If $H$ were a Hamel basis, choose one $b\in H$ and take its rational coefficient homomorphism $c_b$. Its proper measurable kernel $W$ is either positive measure, when F7 gives $W=\mathbb{R}$, or null, when F8 preserves nullness under translation and F6 makes the rational cosets $qb+W$ cover $\mathbb{R}$ by a null set, contradicting the unit interval. For arbitrary additive $f$, the measurable sets $\{x\in [-1,1]:\mathrm{\mid }f(x)\mathrm{\mid }\le n\}$ cover $[-1,1]$; one has positive measure, so F7 bounds $f$ near zero and gives $f(x)=xf(1)$.

For $E\subseteq [0,1{)}^3$ in $K$, the set $A=T^{-1}[E]$ lies in $K$. Step 1.1 supplies Borel $B,N$ with $N$ null and $A\mathrm{△}B\subseteq N$. After intersecting with $D$, bimeasurability and null preservation from step 1.2 give $E\mathrm{△}T[B\cap D]\subseteq T[N\cap D]$, so completeness makes $E$ measurable. Integer translates then cover ${\mathbb{R}}^3$. F9 supplies Countable Choice, exactly the hypothesis of the product and orthogonal-invariance interfaces. Thus every subset of ${\mathbb{R}}^3$ in $K$ is measurable. If a positive-radius closed ball of measure $V$ had a one-use finite partition whose rigid images partitioned two disjoint copies, F8 and finite additivity would give $V=2V$, while inner and outer cubes from F6 give $0<V<\mathrm{\infty }$. A radius-zero ball has one source point and hence one rigid image, not the two target points.

A Vitali selector $V$ would be measurable by step 1.1. If it were null, its explicitly rational-indexed translates would cover $[0,1]$ by a null set; if it had positive measure, arbitrarily many disjoint translates inside $[-1,2]$ would exceed that interval's finite measure. Translation invariance here is F8. For a Bernstein $B$, F1 and F6 make a two-term union of countable sets countable, so one of $B$ and its complement is uncountable; each has no nonempty perfect subset, contradicting step 1.3. If $K$ satisfied AC, F6 would construct a Bernstein set, so full AC fails.

Consequently all stated regularity and anti-choice conclusions hold in $K=L(\mathbb{R})$, without identifying it with $M$ or importing a theorem whose subject is $M$.