Proof. We may assume that $Y = 2^\omega $ and $\mu $ is the usual measure of $2^\omega $ since for every Polish probability space Y, there is a Borel isomorphism between measure $1$ subsets of Y and $2^\omega $ that preserves measure.

Fix a defining formula and a parameter of A. In generic extensions if we write A, we refer to the set defined by the formula and the parameter in the model.

Since A and $\bigcup _{x \in \omega ^\omega } A_x$ are $\boldsymbol {\Sigma }^1_1$ sets, they are Lebesgue measurable. Toward a contradiction, assume that $B := \bigcup _{x \in \omega ^\omega } A_x$ does not have measure 0. Then B has positive measure. By inner regularity of the measure, we can take a closed set $K \subseteq B$ with positive measure. Take a Borel code k of K. We take a random real $r \in 2^\omega $ over V such that $r \in \hat {k}$ .

Now for each $x \in \omega ^\omega \cap V$ , we have $r \not \in A_x$ . In order to prove it, take a Borel code $d_x$ such that $A_x \subseteq \hat {d_x}$ and $\mu (\hat {d_x}) = 0$ . But the condition $A_x \subseteq \hat {d_x}$ is $\boldsymbol {\Pi }^1_1$ . Thus, since the random real avoids $\hat {d_x}$ , we have $r \not \in A_x$ .

Therefore we have

$$\begin{align*}r \not \in \bigcup_{x \in \omega^\omega \cap V} A_x. \end{align*}$$

But in $V[r]$ , it also holds that

$$\begin{align*}(\forall x, x' \in \omega^\omega) (x \le x' \rightarrow A_x \subseteq A_{x'}) \end{align*}$$

since this formula is $\boldsymbol {\Pi }^1_2$ . Thus, by the assumption that $A_x$ is increasing and the fact that the random forcing is $\omega ^\omega $ -bounding, this implies

$$\begin{align*}r \not \in \bigcup_{x \in \omega^\omega} A_x. \end{align*}$$

Therefore, in $V[r]$ , it holds that

$$\begin{align*}(\exists r' \in 2^\omega)(r' \in \hat{k} \smallsetminus B) \end{align*}$$

because $r' = r$ suffices. Recalling B is a $\boldsymbol {\Sigma }^1_1$ set, this statement is written by a $\boldsymbol {\Sigma }^1_2$ formula. Therefore, by Shoenfield’s absoluteness, it also holds in V. That is, there exists an $r' \in 2^\omega $ in V such that

$$\begin{align*}r' \in K \smallsetminus B. \end{align*}$$

This contradicts the choice of K.

Proof. This is proved by the same argument as Theorem 2.1. Recall that $\boldsymbol {\Sigma }^1_3$ - $\mathbb {B}$ -absoluteness follows from $\boldsymbol {\Sigma }^1_2$ measurability (see [Reference Bartoszynski and Judah3, Theorems 9.2.12 and 9.3.8]).

Proof. Assume $A \in \Gamma $ and for each $x \in \omega ^\omega $ , $A_x$ has $\mu $ -measure $0$ and $(\forall x, x' \in \omega ^\omega )(x \le x' \Rightarrow A_{x} \subseteq A_{x'})$ . Put $B_x = \bigcup \{ A_y : x \text { and } y \text { are almost equal} \}$ .

Then by assumption, $B \in \Gamma $ and for each $x \in \omega ^\omega $ , $B_x$ has $\mu $ -measure $0$ and that $(\forall x, x' \in \omega ^\omega )(x \le ^* x' \Rightarrow B_{x} \subseteq B_{x'})$ . Therefore, by $\mathsf {GP}^*(\Gamma )$ , $\bigcup _{x \in \omega ^\omega } B_x$ is a measure $0$ set. Thus $\bigcup _{x \in \omega ^\omega } A_x$ is a measure $0$ set.

Fact 3.1 [Reference Kechris10, Reference Tanaka14]

Let $U \in \Sigma ^1_1$ , $U \subseteq \omega ^\omega \times 2^\omega $ be the universal for $\boldsymbol {\Sigma }^1_1$ subset of $2^\omega $ . Then the relation $\mu (U_x)> r$ for $x \in \omega ^\omega $ and $r \in \mathbb {R}$ is $\Sigma ^1_1$ .

Proof. Take universal sets U and $U^{(2)}$ for $\boldsymbol {\Sigma }^1_1$ subsets of $2^\omega $ and $\omega ^\omega \times 2^\omega $ , respectively, with the following coherent property: $U(S(e, x), y) \iff U^{(2)}(e, x, y)$ , where S is a recursively continuous function. As for the existence of such coherent universal sets, see [Reference Moschovakis11, Section 3.H]. Take $e \in \omega ^\omega $ such that $A(x, y) \iff U^{(2)}(e, x, y)$ . Then we have

$$ \begin{align*} \mu(A_x)> r \iff \mu(U_{S(e, x)}) > r, \end{align*} $$

which is a $\boldsymbol {\Sigma }^1_1$ relation.

Proof. Let $B = (\omega ^\omega \times 2^\omega ) \smallsetminus A$ , which is $\boldsymbol {\Sigma }^1_1$ set. We have

$$ \begin{align*} \mu(A_x) = 0 \iff \mu(B_x) = 1 \iff (\forall n)(\mu(B_x)> 1 - 1/2^n), \end{align*} $$

which is a $\boldsymbol {\Sigma }^1_1$ relation.

Proof. Let $A \subseteq \omega ^\omega \times 2^\omega $ be a $\boldsymbol {\Pi }^1_1$ set. Assume ${\langle A_x : x \in \omega ^\omega \rangle }$ is monotone and each $A_x$ is null. Take a Laver real d over V. $(\forall x \in \omega ^\omega )(\mu (A_x) = 0)$ holds in V and this sentence is $\boldsymbol {\Pi }^1_2$ using Corollary 3.3. So in $V[d]$ , $\mu (A_d) = 0$ holds. Also monotonicity of ${\langle A_x : x \in \omega ^\omega \rangle }$ can be written as a $\boldsymbol {\Pi }^1_2$ formula and holds in V, so it also holds in $V[d]$ . Since d is a dominating real over V, we have $(\bigcup _{x \in \omega ^\omega } A_x)^V \subseteq \bigcup _{x \in \omega ^\omega \cap V} A_x \subseteq A_d$ . Therefore $(\bigcup _{x \in \omega ^\omega } A_x)^V$ is null in $V[d]$ . Since Laver forcing preserves Lebesgue outer measure, it holds that $\bigcup _{x \in \omega ^\omega } A_x$ is null in V.

Proof. In the case of $\mathfrak {b}$ : By assumption, we take an increasing sequence ${\langle A_\alpha : \alpha < \mathfrak {b}\rangle }$ of measure-0 sets such that $\bigcup _{\alpha < \mathfrak {b}} A_\alpha $ doesn’t have measure $0$ . We can take an increasing and unbounded sequence ${\langle x_\alpha : \alpha < \mathfrak {b} \rangle }$ with respect to $\le ^*$ (this sequence is not necessarily cofinal). For each $x \in \omega ^\omega $ , put

$$\begin{align*}\alpha(x) = \min \{ \alpha < \mathfrak{b} : x_\alpha \not \le^* x \}. \end{align*}$$

This is well-defined since ${\langle x_\alpha : \alpha < \mathfrak {b} \rangle }$ is unbounded. And then put

$$\begin{align*}B_x = A_{\alpha(x)}. \end{align*}$$

Now each $B_x$ has measure $0$ and we have

$$ \begin{align*} x \le x' &\Rightarrow x \le^* x' \\ &\Rightarrow (\forall \alpha)(x_\alpha \le^* x \Rightarrow x_\alpha \le^* x') \\ &\Rightarrow \{ \alpha : x_\alpha \not \le^* x' \} \subseteq \{ \alpha : x_\alpha \not \le^* x \} \\ &\Rightarrow \alpha(x) \le \alpha(x') \\ &\Rightarrow B_x \subseteq B_{x'}. \end{align*} $$

Thus ${\langle B_x : x \in \omega ^\omega \rangle }$ is monotone. Also we have $\bigcup _{x \in \omega ^\omega } B_x = \bigcup _{\alpha < \mathfrak {b}} A_\alpha $ . Indeed, it is obvious that the left-hand side is contained in the right-hand side. To prove the reverse inclusion, it is sufficient to show that each $A_\alpha $ is contained in some $B_x$ . So fix $\alpha $ and consider $x = x_{\alpha }$ . Since the sequence ${\langle x_\alpha : \alpha < \mathfrak {b} \rangle }$ is increasing, we have $\alpha \le \alpha (x)$ . Thus $A_\alpha \subseteq A_{\alpha (x)} = B_x$ .

Therefore, $\bigcup _{x \in \omega ^\omega } B_x$ doesn’t have measure $0$ .

In the case of $\mathfrak {d}$ : As above, we can take an increasing sequence ${\langle A_\alpha : \alpha < \mathfrak {d}\rangle }$ of measure-0 sets such that $\bigcup _{\alpha < \mathfrak {d}} A_\alpha $ doesn’t have measure $0$ . By the definition of $\mathfrak {d}$ , we can take a dominating sequence ${\langle x_\alpha : \alpha < \mathfrak {d} \rangle }$ with respect to $\le ^*$ (this sequence is not necessarily increasing).

For each $x \in \omega ^\omega $ , put

$$\begin{align*}\alpha(x) = \min \{ \alpha < \mathfrak{d} : x \le^* x_\alpha \} \end{align*}$$

and put

$$\begin{align*}B_x = A_{\alpha(x)}. \end{align*}$$

One can easily show that ${\langle B_x : x \in \omega ^\omega \rangle }$ is monotone. Also we have $\bigcup _{x \in \omega ^\omega } B_x = \bigcup _{\alpha < \mathfrak {d}} A_\alpha $ . That the left-hand side is contained in the right-hand side is obvious. To show the reverse inclusion, fix $\alpha $ . Since the sequence ${\langle x_\beta : \beta < \alpha \rangle }$ is not a dominating family, we can find an $x \in \omega ^\omega $ such that for all $\beta < \alpha $ , $x \not \le ^* x_\beta $ . Then $\alpha \le \alpha (x)$ . Thus we have $A_\alpha \subseteq A_{\alpha (x)} = B_x$ .

Proof. Clearly there are null towers of length both $\operatorname {add}(\mathsf {null})$ and $\operatorname {non}(\mathsf {null})$ . So using Theorem 4.2, we have this corollary.

Proof. Assume $\operatorname {add}(\mathsf {meager}) = \operatorname {cof}(\mathsf {meager})$ . Let ${\langle M_\alpha : \alpha < \kappa \rangle }$ be a cofinal increasing sequence of meager sets. We can take such a sequence since $\operatorname {add}(\mathsf {meager}) = \operatorname {cof}(\mathsf {meager}) = \kappa $ . For each $\alpha < \kappa $ , take $x_\alpha \in M_{\alpha +1} \smallsetminus M_\alpha $ .

Now recall from Rothberger’s theorem (see, for example, Theorem 5.11 of [Reference Blass4]), there is a Tukey morphism $(\varphi , \psi ) \colon (2^\omega , \mathsf {null}, \in ) \to (\mathsf {meager}, 2^\omega , \not \ni )$ . That is, there are $\varphi \colon 2^\omega \to \mathsf {meager}$ and $\psi \colon 2^\omega \to \mathsf {null}$ such that $\varphi (x) \not \ni y$ implies $x \in \psi (y)$ for every $x, y \in 2^\omega $ .

Using this theorem, we put $N_\alpha = \bigcap _{\beta \ge \alpha } \psi (x_\beta )$ for $\alpha < \kappa $ . Then ${\langle N_\alpha : \alpha < \kappa \rangle }$ is a sequence of null sets of length $\kappa = \mathfrak {b}$ and its union is $2^\omega $ .

Proof. Let G be a $(V, P_\nu )$ -generic filter and work in $V[G]$ . Let ${\langle c_\alpha : \alpha < \operatorname {cf}(\nu ) \rangle }$ be a sequence of Cohen reals added cofinally by $P_\alpha $ .

For a Cohen real c, let $\mathsf {nullset}(c)$ denote the standard null set constructed from c (for a detail, see Definition 3.15 of [Reference Wohofsky15]).

We have the following:

• • For every $x \in \omega ^\omega $ , there is $\alpha < \operatorname {cf}(\nu )$ such that for every $\beta> \alpha $ we have $c_\beta \not <^* x$ .

• • For every $z \in 2^\omega $ , there is $\alpha < \operatorname {cf}(\nu )$ such that for every $\beta> \alpha $ we have $z \in \mathsf {nullset}(c_\beta )$ .

For $x \in \omega ^\omega $ , we let $\alpha _x = \min \{ \alpha : (\forall \beta> \alpha ) (c_\beta \not <^* x) \}$ and let $A_x = \bigcap _{\beta> \alpha _x} \mathsf {nullset}(c_\beta )$ .

We can easily show that each $A_x$ is a null set, the sequence ${\langle A_x : x \in \omega ^\omega \rangle }$ is monotone and the union $\bigcup _{x \in \omega ^\omega } A_x$ is equal to $2^\omega $ . Therefore, ${\langle A_x : x \in \omega ^\omega \rangle }$ is a witness of $\neg \mathsf {GP}(\mathsf {all})$ .

Proof. That (1) implies (2) is shown in Theorem 4.2.

We now prove the converse implication. Assume that $\neg \mathsf {GP}^*(\mathsf {all})$ . Then we can take $A \subseteq \omega ^\omega \times 2^\omega $ such that each section $A_x$ has measure $0$ and $(\forall x, x' \in \omega ^\omega )(x \le ^* x' \Rightarrow A_{x} \subseteq A_{x'})$ holds and $B = \bigcup _{x \in \omega ^\omega } A_x$ does not have measure $0$ . By $\mathfrak {b} = \mathfrak {d} = \kappa $ , we can take a cofinal increasing sequence ${\langle x_\alpha : \alpha < \kappa \rangle }$ with respect to $\le ^*$ . For each $\alpha < \kappa $ , put $C_\alpha = A_{x_\alpha }$ . Then each $C_\alpha $ has measure $0$ . Since $\alpha \mapsto x_\alpha $ is increasing and $x \mapsto A_x$ is monotone, ${\langle C_\alpha : \alpha < \kappa \rangle }$ is also increasing. Also, since ${\langle x_\alpha : \alpha < \kappa \rangle }$ is cofinal, we have $B = \bigcup _{\alpha < \kappa } C_\alpha $ . So $\bigcup _{\alpha < \kappa } C_\alpha $ does not have measure $0$ . Thus ${\langle C_\alpha : \alpha < \kappa \rangle }$ is a null tower of length $\kappa $ .

Proof. This is an example of a reflection argument (see also [Reference Halbeisen8, Chapter 26]).

Take a sequence ${\langle \dot {c}_\beta : \beta < \omega _2\rangle }$ of names of Borel codes such that

$$\begin{align*}\Vdash_{\omega_2} (\forall \beta < \omega_2) (\dot{X}_\beta \subseteq \hat{\dot{c}}_\beta \mathbin{\text{and}} \hat{\dot{c}}_\beta \text{ has measure } 0). \end{align*}$$

For each $\beta < \omega _2$ , take $\gamma _\beta < \omega _2$ such that $\dot {c}_\beta $ is a $P_{\gamma _\beta }$ -name.

Since for each $\alpha < \omega _2$ , $\Vdash _\alpha \mathsf {CH}$ , we can take a sequence ${\langle \dot {x}^\alpha _i : i < \omega _1\rangle }$ such that

$$\begin{align*}\Vdash_\alpha "{\langle\dot{x}^\alpha_i : i < \omega_1\rangle}\text{ is an enumeration of } 2^\omega."\end{align*}$$

For each $\alpha , \beta < \omega _2$ and $i < \omega _1$ , take a maximal antichain $A^{\alpha ,\beta }_i$ such that

$$\begin{align*}A^{\alpha,\beta}_i \subseteq \{p \in P_{\omega_2} : p \Vdash \dot{x}^\alpha_i \in \dot{X}_\beta \mathbin{\text{or}} p \Vdash \dot{x}^\alpha_i \not \in \dot{X}_\beta \}. \end{align*}$$

Since $P_{\omega _2}$ has $\omega _2$ -cc, we can take $\delta ^{\alpha ,\beta }_i < \omega _2$ such that

$$\begin{align*}\bigcup \{ \operatorname{supt}(p) : p \in A^{\alpha,\beta}_i \} \subseteq \delta^{\alpha,\beta}_i. \end{align*}$$

We define a function f from $\omega _2$ into $\omega _2$ as follows:

$$\begin{align*}f(\nu) = \sup \left( \{ \gamma_\beta : \beta \le \nu \} \cup \{ \delta^{\alpha,\beta}_i : \alpha, \beta \le \nu, i < \omega_1 \} \right). \end{align*}$$

Put

$$\begin{align*}C' = \{ \alpha < \omega_2 : \operatorname{cf}(\alpha) = \omega_1, (\forall \nu < \alpha) f(\nu) < \alpha \}. \end{align*}$$

Then clearly $C'$ is $\omega _1$ -club set. So it suffices to show that $C' \subseteq C$ .

Let $\alpha \in C'$ and we shall prove $\alpha \in C$ . Fix $\beta < \alpha $ . Define a $P_\alpha $ -name $\dot {Y}$ by

(*) $$ \begin{align} \Vdash_\alpha "\dot{Y} = \bigcup_{\alpha' < \alpha} \{ \dot{x}^{\alpha'}_i : &(p \Vdash \dot{x}^{\alpha'}_i \in \dot{X}_\beta)^V \nonumber\\ & \text{for some } p \in A^{\alpha',\beta}_i \mathbin{\text{and}} p \upharpoonright \alpha \in \dot{G} \}". \end{align} $$

We claim that $\Vdash _{\omega _2} \dot {X}_\beta \cap V[\dot {G}_\alpha ] = \dot {Y}$ . In order to prove this, take a $(V, P_{\omega _2})$ -generic filter G. In $V[G]$ , take $x \in \dot {X}^G_\beta \cap V[G_\alpha ]$ . Since no new real is added at stage $\alpha $ , we can take $\alpha ' < \alpha $ such that $x \in V[G_{\alpha '}]$ . Thus there is $i < \omega _1$ such that $x = (\dot {x}^{\alpha '}_i)^G$ . Since $(\dot {x}^{\alpha '}_i)^G \in \dot {X}^G_\beta $ , in V, we can take a $p \in G \cap A^{\alpha ',\beta }_i$ such that $p \Vdash \dot {x}^{\alpha '}_i \in \dot {X}_\beta $ . We have $p \in A^{\alpha ',\beta }_i$ . Thus x is an element of $\dot {Y}^G$ .

Conversely, take an element x of $\dot {Y}^G$ . So we can take $\alpha ' < \alpha $ , $i < \omega _1$ and $p \in P_{\omega _2}$ such that

$$\begin{align*}x = (\dot{x}^{\alpha'}_i)^G, (p \Vdash \dot{x}^{\alpha'}_i \in \dot{X}_\beta)^V, p \in A^{\alpha',\beta}_i \mathbin{\text{and}} p \upharpoonright \alpha \in G_\alpha. \end{align*}$$

Clearly we have $x \in V[G_{\alpha '}] \subseteq V[G_\alpha ]$ . Suppose that $(\dot {x}^{\alpha '}_i)^G \not \in \dot {X}^G_\beta $ . Then we can take $q \in G$ such that $q \Vdash \dot {x}^{\alpha '}_i \not \in \dot {X}_\beta $ . By the maximality of $A^{\alpha ',\beta }_i$ , we can take ${r \in A^{\alpha ',\beta }_i \cap G}$ . Since both q and r are elements of G, q and r are compatible. So $r \Vdash \dot {x}^{\alpha '}_i \not \in \dot {X}_\beta $ . Thus p and r are incompatible. But $\operatorname {supt}(p), \operatorname {supt}(r) \subseteq \alpha $ . So $p \upharpoonright \alpha $ and $r \upharpoonright \alpha $ are incompatible. But they are elements of $G_\alpha $ . It contradicts that $G_\alpha $ is a $(V, P_\alpha )$ -generic filter.

Thus we have $\Vdash _{\omega _2} \dot {X}_\beta \cap V[G_\alpha ] \in V[G_\alpha ]$ .

By performing the above operations simultaneously with respect to the $\beta $ , we have

$$\begin{align*}\Vdash_{\omega_2} {\langle\dot{X}_\beta \cap V[\dot{G}_\alpha] : \beta < \alpha \rangle} \in V[\dot{G}_\alpha].\end{align*}$$

Since we have $\Vdash _{\omega _2} \dot {X}_\beta \subseteq \hat {\dot {c}}_\beta $ , it holds that

$$\begin{align*}\Vdash_{\omega_2} "\dot{X}_\beta \cap V[G_\alpha] \subseteq \hat{\dot{c}}_\beta \text{ has measure } 0." \end{align*}$$

Therefore, we have $\alpha \in C$ .

Proof. By Theorem 5.1 and the fact that $ \Vdash _{\omega _2} \mathfrak {b} = \mathfrak {d} = \omega _2$ , it is sufficient to show that

$$\begin{align*}\Vdash_{\omega_2} "\text{There is no null tower of length } \omega_2.\text{"} \end{align*}$$

Let G be a $(V, P_{\omega _2})$ -generic filter. In $V[G]$ , consider an increasing sequence ${\langle A_\alpha : \alpha < \omega _2\rangle }$ of measure $0$ sets. By Lemma 5.2, we can find a stationary set $S \subseteq \omega _2$ such that for all $\alpha \in S$ , $\operatorname {cf}(\alpha ) = \omega _1$ and

$$\begin{align*}({\langle A_\beta \cap V[G_\alpha] : \beta < \alpha \rangle} \in V[G_\alpha] \mathbin{\text{and}} (\forall \beta < \alpha)((A_\beta \cap V[G_\alpha] \text{ has measure } 0)^{V[G_\alpha]}). \end{align*}$$

Fix $\alpha \in S$ . Put $B_\alpha := \bigcup _{\beta < \alpha } A_\beta \cap V[G_\alpha ]$ . Then we have $\bigcup _{\alpha < \omega _2} B_\alpha = \bigcup _{\alpha < \omega _2} A_\alpha $ . We now prove that $B_\alpha $ is also a measure $0$ set in $V[G_\alpha ]$ . Let $\alpha '$ be the successor of $\alpha $ in S. Then $B_\alpha $ is a measure $0$ set in $V[G_{\alpha '}]$ . Since the quotient forcing $P_{\alpha '} / G_\alpha $ is a countable support iteration of the Laver forcing, this forcing preserves positive outer measure. So $B_\alpha $ is also a measure $0$ set in $V[G_{\alpha }]$ .

For each $\alpha \in S$ , take a Borel code $c_\alpha \in \omega ^\omega $ of a measure $0$ set such that $B_\alpha \subseteq \hat {c}_\alpha $ in $V[G_\alpha ]$ . Since $\operatorname {cf}(\alpha ) = \omega _1$ , each $c_\alpha $ appears a prior stage. Then by Fodor’s lemma, we can take a stationary set $S' \subseteq \omega _2$ that is contained in S and $\beta < \omega _2$ such that $(\forall \alpha \in S')(c_\alpha \in V[G_\beta ])$ . But the number of reals in $V[G_\beta ]$ is $\aleph _1$ , so we can take ${T \subseteq S'}$ unbounded in $\omega _2$ and c such that $(\forall \alpha \in T)(c_\alpha = c)$ . Then we have $\bigcup _{\alpha < \omega _2} A_\alpha \subseteq \hat {c}$ in $V[G]$ . So $\bigcup _{\alpha < \omega _2} A_\alpha $ has measure $0$ .

Proof. Assume $\mathsf {CH}$ and let P be the forcing poset from Theorem 5.3, that is, the countable support iteration of Laver forcing notions of length $\omega _2$ . Then we have $P \Vdash \mathsf {GP}(\mathsf {all})$ . In particular we have $P \Vdash \mathsf {GP}(\mathsf {projective})$ . Let $\dot {Q}$ be a P-name of the poset

$$\begin{align*}\operatorname{Fn}(\omega_1, 2, \omega_1) = \{ p : p \text{ is a countable partial function from } \omega_1 \text{ to } 2 \} \end{align*}$$

with the reverse inclusion order. It is well-known that provably $\operatorname {Fn}(\omega _1, 2, \omega _1)$ adds no new reals and forces $\mathsf {CH}$ . So we have $P \ast \dot {Q} \Vdash \mathsf {CH}$ . Since $P \Vdash \mathsf {GP}(\mathsf {projective})$ and $P \Vdash "\dot {Q} \text { adds no new reals}$ ,” we have $P \ast \dot {Q} \Vdash \mathsf {GP}(\mathsf {projective})$ .

Fact 6.1 (Spector–Gandy, see [Reference Chong and Yu5, Proposition 4.4.3])

Let A be a set of reals. Then A is a $\Sigma ^1_2$ set iff there is a $\Sigma _1$ formula $\varphi $ such that

$$\begin{align*}x \in A \iff (L_{\omega_1^{L[x]}}[x], \in) \models \varphi(x). \end{align*}$$

Proof. Let $\mathbf {nBC}$ denote the set of all Borel codes for measure $0$ Borel sets. There is an absolute Tukey morphism $(\varphi , \psi )$ that witnesses $\operatorname {add}(\mathsf {null}) \le \mathfrak {b}$ . That is, $(\varphi , \psi )$ satisfies $\varphi \colon \omega ^\omega \to \mathbf {nBC}$ , $\psi \colon \mathbf {nBC} \to \omega ^\omega $ , and $(\forall x \in \omega ^\omega )(\forall y \in \mathbf {nBC}) (\hat {\varphi (x)} \subseteq \hat {y} \rightarrow x \le ^* \psi (y))$ . By absoluteness, if $x \in \omega ^\omega \cap M$ , then we have $\varphi (x) \in M$ . Now

$$\begin{align*}\bigcup_{x \in \omega^\omega \cap M} \hat{\varphi(x)} \end{align*}$$

has measure 0 since this is contained in $\{ y \in 2^\omega : y \text { is not a random real over } M \}$ by the definition of randomness. Take a $z \in \mathbf {nBC}$ such that

$$\begin{align*}\bigcup_{x \in \omega^\omega \cap M} \hat{\varphi(x)} \subseteq \hat{z}. \end{align*}$$

Now put $w = \psi (z)$ . Then using the fact that $(\varphi , \psi )$ is Tukey morphism, we have w is a dominating real over M.

Proof. Fix a real a. Assume that $L[a] \cap \omega ^\omega $ is unbounded. Note that, in this situation, we have $\omega _1^{L[a]} = \omega _1$ . Let ${\langle x_\alpha : \alpha < \omega _1\rangle }$ be a cofinal increasing sequence in $\omega ^\omega \cap L[a]$ . We can take this sequence with a $\Delta _1(a)$ definition by using a $\Delta _1(a)$ canonical wellordering of $L[a] \cap \omega ^\omega $ . Note that this sequence is unbounded in $V \cap \omega ^\omega $ by assumption.

Take a sequence ${\langle c_\alpha : \alpha < \omega _1\rangle }$ consisting of all Borel codes for measure $0$ Borel sets in $L[a]$ . As above, we can take this sequence with a $\Delta _1(a)$ definition.

For each $x \in \omega ^\omega $ , put

$$\begin{align*}\alpha(x) = \min \{ \alpha : x_\alpha \not \le^* x \}. \end{align*}$$

This is well-defined since ${\langle x_\alpha : \alpha < \omega _1\rangle }$ is unbounded in $V \cap \omega ^\omega $ . Also put

$$\begin{align*}A_x = \bigcup_{\beta < \alpha(x)} \hat{c_\beta}. \end{align*}$$

Then the set A is $\Delta ^1_2(a)$ , by Spector–Gandy theorem and the following equations:

$$ \begin{align*} A &= \{ (x, y) \in \omega^\omega \times 2^\omega : (\exists \beta < \alpha(x))\ y \in \hat{c_\beta} \} \\ &= \{ (x, y) \in \omega^\omega \times 2^\omega : (\exists \alpha)(x_\alpha \not \le^* x \mathbin{\text{and}} (\forall \beta < \alpha)(x_\beta \le^* x) \mathbin{\text{and}} (\exists \beta < \alpha)(y \in \hat{c_\beta})) \} \\ &= \{ (x, y) \in \omega^\omega \times 2^\omega : (\forall \alpha)((x_\alpha \not \le^* x \mathbin{\text{and}} (\forall \beta < \alpha)(x_\beta \le^* x)) \rightarrow (\exists \beta < \alpha)(y \in \hat{c_\beta})) \}. \end{align*} $$

Note that each $A_x$ ( $x \in \omega ^\omega $ ) is a measure $0$ set since it is a countable union of measure $0$ sets. And we can easily observe that $x \le ^* x'$ implies $A_x \subseteq A_{x'}$ .

Since $\alpha \le \alpha (x_\alpha )$ , we have $\bigcup _{x \in \omega ^\omega } A_x = \bigcup _{\alpha < \omega _1} \hat {c_\alpha }$ . Thus it is sufficient to show that $C := \bigcup _{\alpha < \omega _1} \hat {c_\alpha }$ is not a measure $0$ set. In order to show this, assume that C is a measure $0$ set. Note that if a real $y \in 2^\omega $ does not belong to C, then y is a random real over $L[a]$ since the sequence ${\langle c_\alpha : \alpha < \omega _1\rangle }$ enumerates all measure $0$ Borel codes in $L[a]$ . So since we assumed C is measure $0$ , the set $\{ y \in 2^\omega : y \text { is a random real over } L[a] \}$ has measure $1$ . Thus by Lemma 6.2, there is a dominating real over $L[a]$ . This contradicts the assumption.

So we constructed a set A that violates $\mathsf {GP}(\boldsymbol {\Delta }^1_2)$ . This finishes the proof.

Proof. This proof is based on Harrington’s covering game (see also [Reference Moschovakis11, Exercise 6A.17]). In this proof, we use the following notation: for $j < n < \omega $ ,

$$\begin{align*}\operatorname{proj}^n_j : (\omega^\omega)^n \to (\omega^\omega)^{n-1}; (x_0, \dots, x_{n-1}) \mapsto (x_0, \dots, x_{j-1}, x_{j+1}, \dots, x_{n-1}).\end{align*}$$

Fix $B \subseteq \omega ^\omega \times \omega ^\omega \times 2^\omega $ and $A = \operatorname {proj}^3_0(B)$ such that each section $A_x$ has measure $0$ , $(\forall x, x' \in \omega ^\omega )(x \le x' \Rightarrow A_{x} \subseteq A_{x'})$ . Also let $\varepsilon> 0$ . We have to show that the outer measure $\mu ^*(\operatorname {proj}(A))$ is less than or equal to $\varepsilon $ .

Fix a Borel isomorphism $\pi \colon 2^\omega \to \omega ^\omega $ . Consider the following game: At stage n, player I plays $(s_n, t_n, u_n) \in \{0, 1\}^3$ . Player II then plays a finite union $G_n$ of basic open sets such that $\mu (G_n) \le \varepsilon / 16^{n+1}$ . In this game, we define that player I wins if and only if $(z, x, y) \in B$ and $y \not \in \bigcup _{n \in \omega } G_n$ , where $x = \pi (s_0, s_1, \dots ), y = (t_0, t_1, \dots ),$ and $z = \pi (u_0, u_1, \dots )$ .

Assume that player I has a winning strategy $\sigma $ . Put

$$\begin{align*}C &= \{ (z, x, y) \in \omega^\omega \times \omega^\omega \times 2^\omega : (\exists (G_0, G_1, \dots))((z, x, y)\\& \qquad \text{is the play of I along } \sigma \text{ against } (G_0, G_1, \dots)) \}. \end{align*}$$

Then clearly C is a $\boldsymbol {\Sigma }^1_1$ set. Since player I wins, we have $C \subseteq B$ . So we have $\operatorname {proj}^3_0(C) \subseteq \operatorname {proj}^3_0(B) = A$ . So each $(\operatorname {proj}^3_0(C))_x \subseteq A_x$ has measure $0$ . For $x \in \omega ^\omega $ , put $D_x = \bigcup _{x' \le x} (\operatorname {proj}^3_0(C))_{x'}$ , which is a $\boldsymbol {\Sigma }^1_1$ set. Since $(\operatorname {proj}^3_0(C))_x \subseteq A_x$ , each $D_x$ has measure $0$ . And we have $x' \le x$ implies $D_{x'} \subseteq D_x$ .

Thus, by $\mathsf {GP}(\boldsymbol {\Sigma }^1_1)$ , $\operatorname {proj}^2_0(D)$ has measure $0$ . So $\operatorname {proj}^2_0(\operatorname {proj}^3_0(C))$ has also measure $0$ . Therefore we can take $(G_0, G_1, \dots )$ such that $\operatorname {proj}^2_0(\operatorname {proj}^3_0(C)) \subseteq \bigcup _{n \in \omega } G_n$ and $\mu (G_n) \le \varepsilon / 16^{n+1}$ .

Let $(z, x, y)$ be the play along $\sigma $ against $(G_0, G_1, \dots )$ , then $(z, x, y) \in C$ and $y \not \in \bigcup _{n \in \omega } G_n$ . This contradicts to $\operatorname {proj}^2_0(\operatorname {proj}^3_0(C)) \subseteq \bigcup _{n \in \omega } G_n$ .

So player I doesn’t have a winning strategy. Therefore, by $\operatorname {Det}(\Gamma )$ , player II has a winning strategy $\tau $ . Put

$$\begin{align*}E = \bigcup \{ G_n : (G_0, \dots, G_n) \text{ is the play along } \tau \text{ against some } (s_0, t_0, u_0 \dots, s_n, t_n, u_n) \}. \end{align*}$$

Then we have $\operatorname {proj}^2_0(\operatorname {proj}^3_0(B)) \subseteq E$ . In order to check this, let $(z, x, y) \in B$ . Consider the player I’s play $(z, x, y)$ . Let $(G_0, G_1, \dots )$ be the play along $\tau $ against $(z, x, y)$ . Since II wins, $y \in \bigcup _{n \in \omega } G_n \subseteq E$ .

Also we have

$$ \begin{align*} \mu(E) \le \sum_n 8^{n+1} \frac{\varepsilon}{16^{n+1}} = \varepsilon. \end{align*} $$

Therefore we have $\mu ^*(\operatorname {proj}^2_0(A)) \le \mu (E) \le \varepsilon $ .

Fact 8.1.

1. (1) [Reference Bagaria and Woodin2] If $\mathsf {ZFC}$ is consistent, then so is $\mathsf {ZFC}$ plus $\boldsymbol {\Sigma }^1_2$ Lebesgue measurability plus “there is a $\boldsymbol {\Sigma }^1_3$ good wellordering of the reals of length $\omega _1$ .”

2. (2) [Reference Steel13] Assume that there are n many Woodin cardinals. Then there is an inner model $M_n$ of $\mathsf {ZFC}$ that models $\operatorname {Det}(\boldsymbol {\Sigma }^1_n) $ and “ $\text {there is a}\ \boldsymbol {\Sigma }^1_{n+2} \text { good wellordering of the reals}$ .”

Proof. We define a function $G \colon [\omega ^\omega ]^{\le \aleph _0} \to \omega ^\omega $ by

$$\begin{align*}G(S) = \trianglelefteq\text{-minimum } x \text{ such that } x \text{ dominates all elements in } S \end{align*}$$

We define a sequence $(x_\alpha : \alpha < \omega _1)$ of reals in $\omega ^\omega $ by

$$\begin{align*}x_\alpha = G(\{x_\beta : \beta < \alpha \}) \text{ (for } \alpha < \omega_1). \end{align*}$$

Then we put

$$\begin{align*}D = \{ x_\alpha : \alpha < \omega_1 \}. \end{align*}$$

First we claim that D is $\boldsymbol {\Sigma }^1_n$ . Using the usual technique that writes a recursive construction in the way of the existence of an approximation, we have

$$ \begin{align*} x \in D \iff & (\exists \alpha < \omega_1)(\exists F \colon \alpha + 1 \to \omega^\omega) \\ & [(\forall \beta < \alpha)(F(\beta) = G(F \upharpoonright \beta)) \mathbin{\text{and}} x = F(\alpha)]. \end{align*} $$

Eliminating ordinal variables, we have

$$ \begin{align*} x \in D \iff &(\exists z)(\exists w)(\exists f \colon \omega \to \omega^\omega) \\ & \ [w \text{ codes the initial segment below } z \ \mathbin{\text{and}} \\ & \ \ (\forall k)(\exists w') [w' \text{ codes the initial segment below } w(k) \ \mathbin{\text{and}} \\ & \ \ f(k) = G(\{f(i) : \exists j\ ((w)_i = (w')_j) \})] \ \mathbin{\text{and}} \\ & \ x = G(\operatorname{ran} f)]. \end{align*} $$

Therefore D is $\boldsymbol {\Sigma }^1_n$ .

Next, we show that D is $\boldsymbol {\Pi }^1_n$ . As with the above claim, we have

$$ \begin{align*} x \not \in D \iff &(\exists \alpha < \omega_1)(\exists F \colon \alpha + 1 \to \omega^\omega) \\ & [(\forall \beta < \alpha)(F(\beta) = G(F \upharpoonright \beta)) \mathbin{\text{and}} x \le^* F(\alpha) \mathbin{\text{and}} (\forall \beta < \alpha)(x \ne F(\beta))]. \end{align*} $$

By the same transform of formulas in the above claim, we have that $\omega ^\omega \smallsetminus D$ is $\boldsymbol {\Sigma }^1_n$ .

Proof. Let D denote the set defined in Lemma 8.2. We define a set A by

$$\begin{align*}A = \{ (x, y) \in \omega^\omega \times \omega^\omega : y \trianglelefteq z \text{ for the minimum } z \in D \text{ that dominates } x \}. \end{align*}$$

Then we have

$$ \begin{align*} (x, y) \in A \iff & (\exists z)(\exists w) \\ &\ [w \text{ codes the initial segment below } z \mathbin{\text{and}} z \in D \ \mathbin{\text{and}} \\ &\ \ x \le^* z \ \mathbin{\text{and}} (\forall k)((w)_k \in D \rightarrow x \not \le^* (w)_k) \ \mathbin{\text{and}} \\ &\ \ y \trianglelefteq z]. \end{align*} $$

So A is $\boldsymbol {\Sigma }^1_n$ . Moreover, since we have

$$ \begin{align*} (x, y) \in A \iff & (\forall z)(\forall w) \\ &\ [[w \text{ codes the initial segment below } z \mathbin{\text{and}} z \in D \ \mathbin{\text{and}} \\ &\ \ x \le^* z \ \mathbin{\text{and}} (\forall k)((w)_k \in D \rightarrow x \not \le^* (w)_k)] \ \rightarrow \\ &\ \ y \trianglelefteq z], \end{align*} $$

it is also true that A is $\boldsymbol {\Pi }^1_n$ . So A is $\boldsymbol {\Delta }^1_n$ .

Since each $A_x$ is countable and $\bigcup _{x \in \omega ^\omega } A_x = \omega ^\omega $ , this A witnesses $\neg \mathsf {GP}(\boldsymbol {\Delta }^1_n)$ .

Proof. As for (1), combine Fact 8.1 (1), Theorem 2.3, and Lemma 8.3. To show (2), combine Fact 8.1 (2), Theorem 7.1, and Lemma 8.3.

Fact 9.2 (Woodin, see [Reference Bagaria and Bosch1, Lemma 1.2])

If $L(\mathbb {R})^M$ is a Solovay model over V in the weak sense then there is a forcing poset $\mathbb {W}$ in M such that $\mathbb {W}$ adds no new reals and

$$\begin{align*}\mathbb{W} \Vdash "L(\mathbb{R})^M \text{ is a Solovay model over } V \text{ (in the usual sense)}.\text{"} \end{align*}$$

Fact 9.3 [Reference Bagaria and Bosch1, Theorem 2.4]

Suppose that $L(\mathbb {R})^M$ is a Solovay model over V in the weak sense and $\mathbb {P}$ is a strongly- $\boldsymbol {\Sigma }^1_3$ absolutely-ccc poset in M. Let G be an $(M, \mathbb {P})$ -generic filter. Then $L(\mathbb {R})^{M[G]}$ is also a Solovay model in V in the weak sense.

Proof. By Fact 9.2, we may assume that $L(\mathbb {R})^M$ and $L(\mathbb {R})^N$ are Solovay models over V in the usual sense. By universality and homogeneity of the Levy collapse, we have

$$ \begin{align*} M \models "L(\mathbb{R}) \models \varphi(r)" & \iff V[r] \models [\mathrm{Coll}_\kappa \Vdash "L(\mathbb{R}) \models \varphi(r)"] \\ & \iff N \models "L(\mathbb{R}) \models \varphi(r).\text{"}\\[-33pt] \end{align*} $$

Proof. Let $A \subseteq \omega ^\omega \times 2^\omega $ in $L(\mathbb {R})^{V[G]}$ . Take a formula $\varphi $ and an ordinal $\alpha $ such that

$$\begin{align*}A = \{ (x, y) : \varphi(\alpha, x, y) \}^{L(\mathbb{R})^{V[G]}}.\end{align*}$$

In $L(\mathbb {R})^{V[G]}$ , assume that:

1. (1) $(\forall x \in \omega ^\omega ) (\exists c_x \in \omega ^\omega ) (c_x \text { is a Borel code for a measure 0 set} \mathbin {\text {and}} A_x \subseteq \hat {c_x}).$

2. (2) $(\forall x, x' \in \omega ^\omega ) (x \le x' \rightarrow A_x \subseteq A_{x'})$ .

Using the axiom of choice in $V[G]$ we can choose such a family $(c_x : x \in \omega ^\omega )$ . Note that this family is not necessarily in $L(\mathbb {R})^{V[G]}$ .

Since every set of reals is measurable in $L(\mathbb {R})^{V[G]}$ , $\bigcup _{x \in \omega ^\omega } A_x$ is measurable in $L(\mathbb {R})^{V[G]}$ . In order to reach a contradiction, we assume that the measure is positive and take a closed code d in $L(\mathbb {R})^{V[G]}$ such that $\mu (\hat {d})> 0$ and $\hat {d} \subseteq \bigcup _{x \in \omega ^\omega } A_x$ in $L(\mathbb {R})^{V[G]}$ .

Take a random real r over $V[G]$ with $r \in \hat {d}$ . Then by Lemma 9.4, we have in $L(\mathbb {R})^{V[G][r]}$ :

1. (1’) $(\forall x \in \omega ^\omega \cap L(\mathbb {R})^{V[G]}) (A_x \subseteq \hat {c_x})$ , and

2. (2’) $(\forall x, x' \in \omega ^\omega ) (x \le x' \rightarrow A_x \subseteq A_{x'})$ .

By randomness, we have $r \not \in \hat {c_x}$ for all $x \in \omega ^\omega \cap L(\mathbb {R})^{V[G]}$ . But (2’) and the fact that the random forcing is $\omega ^\omega $ -bounding imply $r \not \in A_x$ for all $x \in \omega ^\omega $ in $V[G][r]$ . Thus we have $\hat {d} \smallsetminus \bigcup _{x \in \omega ^\omega } A_x \ne \varnothing $ in $L(\mathbb {R})^{V[G][r]}$ . Then using Lemma 9.4 again, we have $\hat {d} \smallsetminus \bigcup _{x \in \omega ^\omega } A_x \ne \varnothing $ in $L(\mathbb {R})^{V[G]}$ . It is a contradiction to the choice of d.

Fact 10.2 [Reference Blass4, Theorem 2.10]

There is a Borel Tukey connection from $\mathsf {IP}$ to $\omega ^\omega $ . That is, there is Borel maps $\phi \colon \mathsf {IP} \to \omega ^\omega $ and $\psi \colon \omega ^\omega \to \mathsf {IP}$ such that $\phi (I) \le ^* y$ implies $I \sqsubset \psi (y)$ for every $I \in \mathsf {IP}$ and $y \in \omega ^\omega $ .

Proof. Let ${\langle I_n : n \in \omega \rangle }$ be the interval partition with $ \left \lvert {I_n} \right \rvert = n+1$ . Let $A = \{ x \in 2^\omega : (\exists ^\infty n)(x \upharpoonright I_n \equiv 0) \}$ . It is well-known that A is null and comeager. In particular $A \not \in \mathcal {E}$ . For an interval partition $J = {\langle J_k : k \in \omega \rangle }$ , let

$$\begin{align*}A_J = \{ x \in 2^\omega : (\forall^\infty k)(\exists n \in J_k) x \upharpoonright I_n \equiv 0 \}.\end{align*}$$

Each $A_J$ is $F_\sigma $ null set, so it holds that $A_J \in \mathcal {E}$ . And also we have $\bigcup _{J} A_J = A \not \in \mathcal {E}$ .

Moreover, $\{ (J, y) : y \in A_J \}$ is Borel and the family is monotone.

We now have to translate this family to the family indexed by $\omega ^\omega $ .

Let $(\phi , \psi )$ be the Tukey connection in Fact 10.2. Let $B_x = \bigcap _{y \ge ^* x} A_{\psi (y)}$ for $x \in \omega ^\omega $ . ${\langle B_x : x \in \omega ^\omega \rangle }$ is clearly monotone, each $B_x$ is in $\mathcal {E}$ and the family B is $\boldsymbol {\Pi }^1_1$ . On the other hand we have $\bigcup _{J \in \mathsf {IP}} A_J \subseteq \bigcup _{x \in \omega } B_x$ . To see it, let $J \in \mathsf {IP}$ and $a \in A_J$ . Set $x = \phi (J)$ . Then $x \le ^* y$ implies $J \sqsubset \psi (y)$ since $(\phi , \psi )$ is Tukey. So $a \in A_J \subseteq A_{\psi (y)}$ . Therefore we have $a \in B_x$ .

Since $\bigcup _{J \in \mathsf {IP}} A_J \not \in \mathcal {E}$ , we have constructed a counterexample ${\langle B_x : x \in \omega ^\omega \rangle }$ of $\mathsf {GP}(\boldsymbol {\Pi }^1_1, \mathcal {E})$ .