Proof. For (1), this by definition since one can just take $C_n=C$ for all $n\geq 0$ .

For (2), let us identify the countable dense set of Y with natural numbers and let us use variables $i,j$ for this countable dense set, and let d be the metric with respect to which Y is a computable Polish space. Let $r_n>0$ be a computable sequence of reals such that $B(i,r_n)$ is a $\nu $ -computable basis with $B[i,r_n]$ the associated effectively closed ball with the same $\nu $ -measure.Footnote ¹⁵ It suffices to consider the case of U an open ball $B(j,\epsilon )$ where $\epsilon>0$ is rational. In particular it suffices to show that we can compute an index for a c.e. set $I\subseteq \mathbb {N}\times \mathbb {N}$ such that $B(j,\epsilon )=\bigcup _{(i,n)\in I} B[i,r_n]$ . We define $I=\{(i,n)\in \mathbb {N}\times \mathbb {N}: d(i,j)+r_n<\epsilon \}$ . One can then check that one has the identity $B(j,\epsilon )=\bigcup _{(i,n)\in I} B[i,r_n]$ .

In the remainder of the proofs, we often appeal to the fact that if c.e. opens $U_0, U_1, \ldots $ have uniformly computable $\nu $ -measure, then any event in the algebra generated by them also has uniformly computable $\nu $ -measure(cf. [Reference Huttegger, Walsh and Blando15, Proposition 2.8(2)]).

For (3), it suffices to show it for a single intersection, and in this case just consider the intersection $C_n\cap C_n^{\prime }$ where $C_n$ witnesses that B is $\Sigma ^{0,\nu }_2$ and $C_n^{\prime }$ witnesses that $B^{\prime }$ is $\Sigma ^{0,\nu }_2$ .

For (4), since V is $\Sigma ^{0,\nu }_2$ , let $V=\bigcup _n D_n$ , where $D_n$ is effectively closed with $\nu (D_n)$ uniformly computable. Without loss of generality $D_n$ is increasing. Hence, using computability of $\nu (V)$ , we can compute n such that $\nu (V\setminus D_n)<\epsilon $ .

For (5) simply apply (4) to the relative complement.

Proof. Let $B=\bigcup _n C_n$ where $C_n$ is uniformly effectively closed and $\nu (C_n)$ is computable; and where in addition $\nu (B)$ is computable. Without loss of generality $C_n$ is increasing. Define uniformly computable reals $0\leq \eta _n\leq 1$ by $\eta _0=\nu (C_0)$ and $\eta _n=\nu (C_n\setminus C_{n-1})$ for $n>0$ . Since $\sum _n \eta _n = \nu (B)$ , we have $\lim _n \eta _n=0$ . Further $\nu (B\setminus C_m)=\sum _{n>m} \eta _n$ . Let $\delta _n$ be a uniformly computable sequence of rationals with $0\leq \delta _n-\eta _n<2^{-(n+1)}$ . Since $\delta _n<2^{-(n+1)}+\eta _n$ , also $\lim _n \delta _n=0$ . Further:

(2.1) $$ \begin{align} \sum_{n>m} \delta_n=\sum_{n>m} \eta_n+\sum_{n>m} (\delta_n-\eta_n)<\nu(B\setminus C_m)+2^{-(m+1)}. \end{align} $$

By Proposition 2.3(5), uniformly compute c.e. open $U_n\supseteq C_n$ with $\nu (U_n)$ computable and $\nu (U_n\setminus C_n)<\frac {\epsilon }{2}\cdot \delta _n$ . Let $U=\bigcup _n U_n$ . This gives the required estimate

$$ \begin{align*} \nu(U\setminus B)\leq \sum_n \nu(U_n\setminus C_n)< \sum_n \frac{\epsilon}{2}\cdot \delta_n \leq \sum_n \frac{\epsilon}{2}\cdot \eta_n + \sum_n \frac{\epsilon}{2} \cdot (\delta_n-\eta_n)\leq \epsilon. \end{align*} $$

It remains to show that $\nu (U)$ is computable. Since $\nu (B)$ is computable and $\nu (U)=\nu (B)+\nu (U\setminus B)$ , it suffices to show that $\nu (U\setminus B)$ is computable. For $n>0$ :

(2.2) $$ \begin{align} \nu(U_n\setminus C_{n-1})\leq \nu(U_n\setminus C_n)+\nu(C_n\setminus C_{n-1})\leq \epsilon\cdot \delta_n+ \delta_n = (1+\epsilon)\cdot \delta_n. \end{align} $$

For $m>0$ , note that one has the inclusions

(2.3) $$ \begin{align} (\bigcup_{n\geq m} U_n)\setminus C_m\subseteq (U_m\setminus C_m)\cup \bigcup_{n>m} (U_n\setminus U_{n-1})\subseteq (U_m\setminus C_m) \cup \bigcup_{n>m} (U_n\setminus C_{n-1}), \end{align} $$

where the last inclusion follows from $C_{n-1}\subseteq U_{n-1}$ . Hence for all $m> 0$ one has that $\nu (U\setminus C_m)-\nu ((\bigcup _{n< m} U_n)\setminus C_m)$ is

$$ \begin{align*} \leq \;& \nu((\bigcup_{n\geq m} U_n)\setminus C_m) \leq \nu(U_m\setminus C_m)+\sum_{n>m} \nu(U_n\setminus C_{n-1}) \\ \leq \;& \nu(U_m\setminus C_m)+\sum_{n>m} (1+\epsilon)\cdot \delta_n < \epsilon\cdot \delta_m+(1+\epsilon)\cdot \nu(B\setminus C_m) +(1+\epsilon)\cdot 2^{-(m+1)}, \end{align*} $$

where in the first line we apply (2.3) and where in the last line we apply (2.2) and then (2.1). Then for all $m> 0$ we have that $\left |\nu (U\setminus B)-\nu ((\bigcup _{n<m} U_n)\setminus C_m)\right |$ is

$$ \begin{align*} & \leq \left|\nu(U\setminus B)-\nu(U\setminus C_m)\right|+\left| \nu(U\setminus C_m)-\nu((\bigcup_{n< m} U_n)\setminus C_m)\right| \\&\leq \nu(B\setminus C_m)+\epsilon\cdot \delta_m+(1+\epsilon)\cdot \nu(B\setminus C_m) +(1+\epsilon)\cdot 2^{-(m+1)}. \end{align*} $$

Since the quantity in this last line is uniformly computable and goes to zero as m goes to infinity, we can compute a subsequence which witnesses that $\nu (U\setminus B)$ is a computable real. We appeal here to the fact that $(\bigcup _{n<m} U_n)\setminus C_m$ has computable $\nu $ -measure, since it is in the algebra generated by the uniformly computable sequence $U_n, Y\setminus C_n$ of c.e. opens with $\nu $ -computable measure (cf. [Reference Huttegger, Walsh and Blando15, Proposition 2.8(2)]).

Proof. For the forward direction, suppose that $f_n$ and C are as specified; we want to show that $\lim _n f_n=0$ on $\mathsf {SR}^{\nu }\cap C$ . Suppose that $\epsilon>0$ is rational. For each $m\geq 0$ , uniformly compute a real $\epsilon _{m}$ in the interval $(\frac {\epsilon }{2}, \epsilon )$ such that the c.e. open $\{y: \sup _{n\geq m} f_n(y)>\epsilon _{m}\}$ has uniformly computable $\nu $ -measure.Footnote ¹⁶ Then its relative complement $D_{m}: = \{y: \sup _{n\geq m} f_n(y)\leq \epsilon _{m}\}$ is effectively closed and has uniformly computable $\nu $ -measure as well. Finally, using a $\nu $ -computable basis, let $Y\setminus C=\bigcup _m U_m$ , where $U_m$ is a computable sequence of c.e. opens from the $\nu $ -computable basis, and let $C_m$ be effectively closed superset of $U_m$ with the same $\nu $ -measure. Consider the class $V_{\epsilon } = (\bigcup _{m} (D_{m}\cap C)) \cup (\bigcup _m C_m)$ , which is an effective union of effectively closed classes each with uniformly computable measure. Since by hypothesis $f_n\rightarrow 0\, \nu $ -a.s. on C, there are $\nu (C)$ -many y in C such that there is $m\geq 0$ such that for all $n\geq m$ one has $f_n(y)<\frac {\epsilon }{2}$ , and so $\sup _{n\geq m} f_n(y)\leq \frac {\epsilon }{2}<\epsilon _{m}$ and so this y would be in $D_{m}$ . Hence $\nu (V_{\epsilon })=1$ , so that $V_{\epsilon }$ is a $\Sigma ^{0,\nu }_2$ class of $\nu $ -measure one. Let y in $\mathsf {SR}^{\nu }\cap C$ . Then by Proposition 1.5, y is in $V_{\epsilon }$ . But since y is in $\mathsf {KR}^{\nu }\cap C$ , we cannot have that y is in $\bigcup _m C_m$ . (For suppose that y is $C_m$ for some $m\geq 0$ . Since $C_m$ is an effectively closed superset of c.e. open $U_m$ with the same $\nu $ -measure, it follows that $C_m\setminus U_m$ is effectively closed and $\nu (C_m\setminus U_m)=0$ . Then since y is in $\mathsf {KR}^{\nu }$ , we have that y is not in $C_m\setminus U_m$ . Hence y is in $U_m$ as well. But $U_m$ is by definition a subset of $Y \setminus C$ ). Thus rather we must have that y is in $D_{m}\cap C$ for some $m\geq 0$ , and so $\sup _{n\geq m} f_n(y)\leq \epsilon _{m}<\epsilon $ .

The backward direction is easy since if $U_n$ is a decreasing sequential Schnorr $\nu $ -test and we set $f_n=I_{U_n}$ , then $\sup _{n\geq m} f_n =I_{\bigcup _{n\geq m} U_n}$ is an $L_1(\nu )$ Schnorr test.

Proof. For (1), suppose that $U=\bigcup _n V_n$ , where $V_n$ comes from a $\nu $ -computable basis. Then $\|I_U-I_{\bigcup _{m<n} V_m}\|_1 =\nu (U)-\nu (\bigcup _{m<n} V_m)$ is computable and goes to zero. Then we can compute sequence $n_{\ell }$ such that $\|I_U-I_{\bigcup _{m<n_{\ell }} V_m}\|_1\leq 2^{-\ell }$ . Hence $I_U$ is $L_1(\nu )$ computable since $I_{\bigcup _{m<n_{\ell }} V_m}$ comes from the countable dense set. The proof of (2) follows by taking complements; the proof of (3) is similar to that of (1).

Proof. For (1), the three $\nu $ -a.s. non-negative $L_1(\nu )$ computable functions $f,g,h$ are such that their pushforwards $f\#\nu , g\#\nu , h\#\nu $ are computable probability measures on $\mathbb {R}^{\geq 0}$ . Using the Effective Baire Category Theorem, we can compute a uniformly computable sequence of reals $r_i$ , dense in $\mathbb {R}^{>0}$ , such that the events $f^{-1}(r_i,\infty ]$ and $g^{-1}(r_i,\infty ]$ and $h^{-1}(r_i,\infty ]$ have uniformly computable $\nu $ -measure.Footnote ¹⁷ By doing this again for the non-negative $L_1(\nu )$ computable functions $f+r_i,g+r_i,h+r_i$ , we can compute a uniformly computable sequence of reals $s_k$ , dense in $\mathbb {R}^{>0}$ , such that the events $(f+r_i)^{-1}(s_k,\infty ]$ and $(g+r_i)^{-1}(s_k,\infty ]$ and $(h+r_i)^{-1}(s_k,\infty ]$ have computable $\nu $ -measure, uniformly in $i,k\geq 0$ . Define c.e. index set $I=\{(j,k): s_k<r_j\}$ . For $i\geq 0$ and $(j,k)$ in I, define $\Sigma ^{0,\nu }_2$ class $B_{i,j,k}=g^{-1}(r_j,\infty ]\cap (h+r_i)^{-1}[0,s_k]$ , and then define $B_i:=\bigcup _{(j,k)\in I} B_{i,j,k}$ . Then on $\mathsf {SR}^{\nu }$ we have the identity $f^{-1}(r_i,\infty ]=B_i$ . Since $f^{-1}(r_i,\infty ]$ has $\nu $ -computable measure, so does $B_i$ , and hence it too is a $\Sigma ^{0,\nu }_2$ class by Proposition 2.3(3). Then we can finish by using the density of the $r_i$ sequence on $\mathbb {R}^{>0}$ to compute a subsequence $a_n<r_{i_n}<b_n$ .

For (2), fix $s>0$ to be rational. Let $b_n=s$ and let $a_n$ converge up to s. Let $g_n=g+s, h_n=h+s, f_n=f+s$ . Apply part (1) to $g_n, h_n, f_n$ and $(a_n, b_n)$ to get a computable sequence $r_n$ and $B_n$ . Then $f\geq 0\, \nu $ -a.s. implies $f_n\geq s\, \nu $ -a.s. which implies that $f_n^{-1}(r_n, \infty ]$ has $\nu $ -measure one, which implies that $B_n$ has $\nu $ -measure one. Then each $B_n$ has $\nu $ -measure one, and so $\mathsf {SR}^{\nu }$ is a subset of each $B_n$ by Proposition 1.5. Then $f_n\geq r_n\geq a_n$ on $\mathsf {SR}^{\nu }$ for all $n\geq 0$ , and so $f+s\geq s$ on on $\mathsf {SR}^{\nu }$ and hence $f\geq 0$ on $\mathsf {SR}^{\nu }$ .

For (3), the argument is just like (1) but with the inequalities reversed.

Proof. We must show that for c.e. open $W\subseteq S$ we have that $\mu (W)$ is left-c.e. We can write $W=\bigcup _i (U_i\times [\sigma _i])$ , where $U_i\subseteq \Omega $ is a computable sequence from a p-computable basis with effectively closed $C_i\supseteq U_i$ with the same p-measure, and $\sigma _i$ is a computable sequence from T. It suffices to show that for all n, we have that $W_n=\bigcup _{i<n} (U_i\times [\sigma _i])$ has uniformly left-c.e. $\mu $ -measure. For this, in turn, it suffices to show the same of $W_{n,i}:=\big (U_i\times [\sigma _i]\big )\setminus \bigcup _{j<i} \big ( U_j\times [\sigma _j] \big )$ for each $i<n$ .

Fix $i<n$ and for $j<i$ , define computable set $I_{i,j}\subseteq T$ so that $[\sigma _i]\setminus [\sigma _j]=\bigcup _{\eta \in I_{i,j}} [\eta ]$ . Since $I_{i,j}$ is computable let $I_{i,j,s}$ be a computable non-decreasing sequence such that $I_{i,j}=\bigcup _s I_{i,j,s}$ and $J_{i,j,s}:=I_{i,j,s+1}\setminus I_{i,j,s}$ has at most one element.

For each $\tau $ in $2^{<\mathbb {N}}$ with length $i>0$ and each $\varsigma $ in $\mathbb {N}^{<\mathbb {N}}$ of length i, define the following c.e. open $U_{\tau }\subseteq \Omega $ and c.e. open $V_{\tau ,\varsigma }\subseteq \mathcal {X}$ , wherein j ranges over $<i$ :

$$ \begin{align*} U_{\tau} = \big(\bigcap_{\tau(j)=1} U_i\setminus C_j\big) \cap \big(\bigcap_{\tau(j)=0} U_i\cap U_j\big), \hspace{2mm} V_{\tau,\varsigma} = \big(\bigcap_{\tau(j)=1} [\sigma_i]\big) \cap \big(\bigcap_{\tau(j)=0, \eta\in J_{i,j,\varsigma(j)}} [\eta]\big). \end{align*} $$

Let $I_i$ be the computable index set ${\{(\tau , \varsigma ){\kern-1pt}\in{\kern-1pt} 2^i{\kern-1pt}\times{\kern-1pt} \mathbb {N}^i: \forall \; j<i \; (\tau (j){\kern-1pt}={\kern-1pt}1\rightarrow \varsigma (j){\kern-1pt}={\kern-1pt}0)\}}$ . Then on $\mathsf {KR}^{p}\times \mathcal {X}$ one has $W_{n,i}=\bigsqcup _{(\tau ,\varsigma )\in I_i} (U_{\tau }\times V_{\tau ,\varsigma })$ , where the disjoint union is disjoint only on $\mathsf {KR}^{p}\times \mathcal {X}$ . Then $\mu (W_{n,i})=\sum _{(\tau ,\varsigma )\in I_i} \mu (U_{\tau }\times V_{\tau ,\varsigma })$ .

Hence, we have reduced the problem to showing that $\mu (U\times [\sigma ])$ is left-c.e. when $U\subseteq \Omega $ is c.e. open. Let $U=\bigcup _i U_i$ , where $U_i$ is an increasing sequence of c.e. opens with uniformly p-computable measure. By Definition 1.1(3), one has that $\theta \mapsto p(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )$ is uniformly $L_1(p)$ computable, and hence so is $\theta \mapsto I_{U_i}(\theta )\cdot p(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )$ by Propositions 2.6(1) and 2.7. Then $\mu (U\times [\sigma ])>q$ iff there is $i\geq 0$ with $\mu (U_i\times [\sigma ])>q$ , which by definition happens iff $\int _{U_i} p(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta ) \; dp(\theta )>q$ . Since $\int _{U_i} p(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta ) \; dp(\theta )$ is a computable real uniformly in $i\geq 0$ , we are done and indeed $\mu (U\times [\sigma ])$ is left-c.e.

Proof. By the same considerations as at the end of the last proof.

Proof. We suppose that there is a uniformly computable sequence of $L_1(p)$ Schnorr tests $g(\sigma \hspace {-1mm}\mid \hspace {-1mm} \cdot ), h(\sigma \hspace {-1mm}\mid \hspace {-1mm} \cdot )$ such that $p(\sigma \hspace {-1mm}\mid \hspace {-1mm} \cdot )=g(\sigma \hspace {-1mm}\mid \hspace {-1mm} \cdot )-h(\sigma \hspace {-1mm}\mid \hspace {-1mm} \cdot )$ pointwise. By Proposition 2.8(2) we have $g(\sigma \hspace {-1mm}\mid \hspace {-1mm} \cdot )\geq h(\sigma \hspace {-1mm}\mid \hspace {-1mm} \cdot )$ on $\mathsf {SR}^{p}$ .

Where $\emptyset $ is the root of the tree T, we show that $p(\emptyset \hspace {-1mm}\mid \hspace {-1mm} \theta )=1$ for $\theta $ in $\mathsf {SR}^{p}$ :

1. – Suppose $0<s<1$ is rational. By Proposition 2.8, choose a computable r in $(s,1)$ , and a $\Sigma ^{0,p}_2$ set B which is equal to $\{\theta : p(\emptyset \hspace {-1mm}\mid \hspace {-1mm} \theta )>r\}$ on $\mathsf {SR}^{p}$ . Then $p(\emptyset \hspace {-1mm}\mid \hspace {-1mm} \theta )> r>s$ for $\theta $ in $\mathsf {SR}^{p}$ by Proposition 1.5. Since this holds for all rational s in $(0,1)$ we have $p(\emptyset \hspace {-1mm}\mid \hspace {-1mm} \theta )\geq 1$ for $\theta $ in $\mathsf {SR}^{p}$ .

2. – Suppose $s>1$ is rational. By Proposition 2.8, choose a computable r in $(1,s)$ and a $\Sigma ^{0,p}_2$ set $B^{\prime }$ which is equal to $\{\theta : p(\emptyset \hspace {-1mm}\mid \hspace {-1mm} \theta )>r\}$ on $\mathsf {SR}^{p}$ . But then $B^{\prime }$ is p-null, and no element of $\mathsf {KR}^{p}$ can be in a $\Sigma ^{0,p}_2\, p$ -null set. Since this holds for all rational $s>1$ , we have $p(\emptyset \hspace {-1mm}\mid \hspace {-1mm} \theta )\leq 1$ for $\theta $ in $\mathsf {SR}^{p}$ .

Suppose that $\sigma $ in T is fixed; enumerate its immediate successors in T as $\tau _0, \tau _1, \ldots $ . We may suppose by induction that we know $0\leq p(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )\leq 1$ on $\theta $ in $\mathsf {SR}^{p}$ . We want to show that $p(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )=\sum _i p(\tau _i \hspace {-1mm}\mid \hspace {-1mm} \theta )$ for all $\theta $ in $\mathsf {SR}^{p}$ :

1. – First we show that for all $\theta $ in $\mathsf {SR}^{p}$ , we have $p(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )\leq \sum _i p(\tau _i \hspace {-1mm}\mid \hspace {-1mm} \theta )$ . By Proposition 2.8, choose dense computable $r_n$ sequence in $(0,1)$ and a computable sequence of $\Sigma ^{0,p}_2$ sets $B_{n,j}$ which are equal to $\{\theta : p(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )> r_n> \sum _{i\leq j} p(\tau _i \hspace {-1mm}\mid \hspace {-1mm} \theta )\}$ on $\mathsf {SR}^{p}$ . Fix $n\geq 0$ and set $B_n:=\bigcap _j B_{n,j}$ . Since $B_n$ is p-null by our standing effective assumptions (in particular by the second conjunct of Definition 1.1(3)), we can compute a subsequence $B_{n,j_m}$ with $\nu (B_{n,j_m})\leq 2^{-m}$ for all $m\geq 0$ . By Proposition 1.4, no element of $\mathsf {SR}^{p}$ is in $B_n$ . Since this holds for all $n\geq 0$ and since $r_n$ is dense in $(0,1)$ , we are done.

2. – Second we show that for all $\theta $ in $\mathsf {SR}^{p}$ , we have $p(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )\geq \sum _i p(\tau _i \hspace {-1mm}\mid \hspace {-1mm} \theta )$ . Fix ${j>0}$ . By Proposition 2.8, choose dense computable $r_n$ sequence in $(0,1)$ and a computable sequence of $\Sigma ^{0,p}_2$ sets $B_n$ which are equal to $\{\theta : p(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )< r_n< \sum _{i\leq j} p(\tau _i \hspace {-1mm}\mid \hspace {-1mm} \theta )\}$ on $\mathsf {SR}^{p}$ . But $B:=\bigcup _n B_n$ is p-null by our standing effective assumptions (in particular by the second conjunct of Definition 1.1(3)), and hence by Proposition 2.3(3), B is a $\Sigma ^{0,p}_2\, p$ -null class. No element of $\mathsf {KR}^{p}$ can be in a $\Sigma ^{0,p}_2\, p$ -null set. Since this holds for all $j>0$ , we have that $p(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )\geq \sum _i p(\tau _i \hspace {-1mm}\mid \hspace {-1mm} \theta )$ for all $\theta $ in $\mathsf {SR}^{p}$ .

Proof. For rational $q>0$ , one has $p(U \hspace {-1mm}\mid \hspace {-1mm} x_{n})>q$ iff there is $\sigma $ in T of length n such that x is in $[\sigma ]$ and there is rational $r>0$ with $\mu (U\times [\sigma ])>r$ and ${0<\mu (\Omega \times [\sigma ])\leq \frac {r}{q}}$ . This is a c.e. open condition in variable x, since $U\times [\sigma ]$ is c.e. open in S and since $\Omega \times [\sigma ]$ is both c.e. open and effectively closed in S.

Proof. We first prove that (1) and (2) hold for c.e. opens V in S which are finite unions $\bigcup _{i\leq n} (W_i\times [\sigma _i])$ of elements from the $\mu $ -computable basis. For each $i\leq n$ , choose $C_i\supseteq W_i$ effectively closed with the same p-measure as $W_i$ . For each $J\subseteq [0,n]$ let $J^{\prime }=\{i\in J: \forall \; j\in J\; \sigma _j\nprec \sigma _i\}$ and define the following events where i in the subscript ranges over $\leq n$ :

$$ \begin{align*} A_J=\bigcap_{i\in J} W_i\cap \bigcap_{i\notin J} \Omega\setminus W_i, \hspace{5mm} B_J=\bigcap_{i\in J} W_i\cap \bigcap_{i\notin J} \Omega\setminus C_i. \end{align*} $$

Then $p(V^{\theta } \hspace {-1mm}\mid \hspace {-1mm} \theta )=\sum _{J\subseteq [0,n]} \sum _{i\in J^{\prime }} I_{A_J}(\theta ) \cdot p(\sigma _i \hspace {-1mm}\mid \hspace {-1mm} \theta )$ . By Propositions 2.6 and 2.7 this is a finite sum of $L_1(p)$ computable functions, since $A_J$ comes from the algebra generated by the p-computable basis and since $\theta \mapsto p(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )$ is $L_1(p)$ computable uniformly in $\sigma $ from T. This takes care of (1). For (2), it suffices to find one such version, since all such versions agree on the Schnorr randoms. Suppose that $p(\sigma \hspace {-1mm}\mid \hspace {-1mm} \cdot )$ uses the version which is the difference of $g(\sigma \hspace {-1mm}\mid \hspace {-1mm} \cdot )$ and $ h(\sigma \hspace {-1mm}\mid \hspace {-1mm} \cdot )$ , where these are $L_1(p)$ Schnorr tests, uniformly in $\sigma $ from T. Then to finish (2), we note that $p(V^{\theta } \hspace {-1mm}\mid \hspace {-1mm} \theta )$ has a version $f(V \hspace {-1mm}\mid \hspace {-1mm} \theta )$ which is the difference of

(3.1) $$ \begin{align} g(V \hspace{-1mm}\mid\hspace{-1mm} \theta):=\sum_{J\subseteq [0,n]} \sum_{i\in J^{\prime}} I_{B_J}(\theta) \cdot g(\sigma_i \hspace{-1mm}\mid\hspace{-1mm} \theta), \hspace{5mm} h(V \hspace{-1mm}\mid\hspace{-1mm} \theta):=\sum_{J\subseteq [0,n]} \sum_{i\in J^{\prime}} I_{B_J}(\theta) \cdot h(\sigma_i \hspace{-1mm}\mid\hspace{-1mm} \theta). \end{align} $$

Second, we treat the general case. Let $U=\bigcup _m U_m$ , where $U_m$ comes from the $\mu $ -computable basis. Let $V_{\ell }=\bigcup _{m\leq \ell } U_{m}$ . For each $n\geq 0$ , compute $\ell _n\geq 0$ such that $\mu (U\setminus V_{\ell _n})<2^{-n}$ . One has that $\| p(U^{\cdot } \hspace {-1mm}\mid \hspace {-1mm} \cdot ) - p(V_{\ell _n}^{\cdot } \hspace {-1mm}\mid \hspace {-1mm} \cdot ) \|_{L_1(p)}$ is equal to

$$ \begin{align*} \int_{\Omega} \int_{\mathcal{X}} I_U(\theta,x)-I_{V_{\ell_n}}(\theta,x) \; dp(\cdot \hspace{-1mm}\mid\hspace{-1mm} {\theta})(x) \; dp(\theta) = \mu(U\setminus V_{\ell_n})< 2^{-n}. \end{align*} $$

By (1) applied to $V_{\ell _n}$ , we have that $\theta \mapsto p(V_{\ell _n} \hspace {-1mm}\mid \hspace {-1mm} \theta )$ is $L_1(p)$ computable; and we just showed that these converge fast to $\theta \mapsto p(U \hspace {-1mm}\mid \hspace {-1mm} \theta )$ in $L_1(p)$ ; and so we are done with (1). For (2), let $\theta \mapsto p(U^{\theta } \hspace {-1mm}\mid \hspace {-1mm} \theta )$ have a version f which is a difference of $L_1(p)$ Schnorr tests $g,h$ . By (2) applied to $V_i$ , we have that $f(V_i \hspace {-1mm}\mid \hspace {-1mm} \theta )$ is increasing and its limit exists in $[0,1]$ , where $f(V_i \hspace {-1mm}\mid \hspace {-1mm} \theta )$ is the difference of $g(V_i \hspace {-1mm}\mid \hspace {-1mm} \theta ), h(V_i \hspace {-1mm}\mid \hspace {-1mm} \theta )$ from (3.1). It suffices to show the identity $f(\theta )=\lim _i f(V_i \hspace {-1mm}\mid \hspace {-1mm} \theta )$ for $\theta $ in $\mathsf {SR}^{p}$ . The proof of this identity, via the two inequalities, follows as in the proof of Proposition 3.5.

Proof. Fix $\theta $ in $\mathsf {SR}^{p}$ . Suppose $(\theta ,x)$ is not in $\mathsf {SR}^{\mu }$ . Choose a sequential Schnorr $\mu $ -test $U_n$ with $(\theta ,x)$ in $\bigcap _n U_n$ . By Proposition 3.8 the map $\theta ^{\prime }\mapsto p(U_n^{\theta ^{\prime }} \hspace {-1mm}\mid \hspace {-1mm} \theta ^{\prime })$ is $L_1(p)$ computable, uniformly in $n\geq 0$ . By Proposition 2.8, there is a computable sequence of computable reals $\eta _n$ in the interval $(2^{-n}, 2^{-n+1})$ and a computable sequence of $\Sigma ^{0,p}_2$ classes $B_n$ such that $B_n=\{\theta ^{\prime }: p(U_{2n}^{\theta ^{\prime }} \hspace {-1mm}\mid \hspace {-1mm} \theta ^{\prime })>\eta _n\}$ on $\mathsf {SR}^{p}$ . Then:

$$ \begin{align*} 2^{-2n}\geq \mu(U_{2n}) = \int_{\Omega} p(U_{2n}^{\theta^{\prime}} \hspace{-1mm}\mid\hspace{-1mm} \theta^{\prime}) \; dp(\theta^{\prime}) \geq \int_{B_n} p(U_{2n}^{\theta^{\prime}} \hspace{-1mm}\mid\hspace{-1mm} \theta^{\prime}) \; dp(\theta^{\prime}) \geq 2^{-n} p(B_n). \end{align*} $$

By Proposition 1.4 and by $\theta $ being in $\mathsf {SR}^{p}$ , there is $n_0> 0$ such that for all $n\geq n_0$ one has $p(U_{2n}^{\theta } \hspace {-1mm}\mid \hspace {-1mm} \theta ) \leq 2^{-n+1}$ . Then by Proposition 3.8(2), the sample x is in the $p(\cdot \hspace {-1mm}\mid \hspace {-1mm} {\theta })$ -null set $\bigcap _{n\geq n_0} U_{2n}^{\theta }$ . With $\theta $ fixed, there are only countably many such null sets for each Schnorr $\mu $ -test $U_n$ . And since there are only countably many Schnorr $\mu $ -tests, there are only $p(\cdot \hspace {-1mm}\mid \hspace {-1mm} {\theta })$ -null many x such that $(\theta ,x)$ is not in $\mathsf {SR}^{\mu }$ .

Proof. This follows from the identity $\mu (\Omega \times [\sigma ] \mid [\theta _n]\times \mathcal {X})=\frac {1}{p([\theta _n])} \int _{[\theta _n]} p(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta ^{\prime }) dp(\theta ^{\prime })$ and the effective version of Lévy’s Upward Theorem ([Reference Huttegger, Walsh and Blando15]; see discussion therein of [Reference Pathak, Rojas and Simpson20, Reference Rute22]).

Proof. First suppose that $V\subseteq \Omega $ is an element of the p-computable basis, with effectively closed $C\supseteq V$ such that $V,C$ have same p measure. We first show that $I_{f^{-1}(V)}$ is a computable point of $L_1(\mu _{X})$ .

Let $D\subseteq \Omega \times \mathcal {X}$ be a $\mu $ -measure one subset of $\mathsf {KR}^{p}\times \mathcal {X}$ such that for $(\theta , x)$ in D we have $\lim _n f_n(x)$ exists and is equal to $\theta $ (cf. remarks immediately after Proposition 3.4).

Let us show that $\lim _n (I_{f^{-1}_n(V)}\circ X)=I_{f^{-1}(V)}\circ X\, \mu $ -a.s. It suffices to show that it happens on D. Suppose that $(\theta , x)$ is in D.

1. – First suppose that $I_{f^{-1}(V)}(x)=1$ . Then $f(x)$ in V. Since V is open and we are on D, eventually $f_n(x)$ in V, which implies that eventually $I_{f^{-1}_n(V)}(x)=1$ .

2. – Second suppose that $I_{f^{-1}(V)}(x)=0$ . Then $f(x)$ is in $\Omega \setminus V$ . Since we are on $D\subseteq \mathsf {KR}^{p}\times \mathcal {X}$ , we have that $\theta =f(x)$ is in the c.e. open $\Omega \setminus C$ . Suppose that there were infinitely many $n\geq 0$ with $I_{f_n^{-1}(V)}(x)=1$ , so that there were infinitely many $n\geq 0$ with $f_n(x)$ in V. Since we are on D, we then have that $f(x)$ is in the topological closure $\overline {V}$ of V, and hence $f(x)$ in C, since C is an effectively closed superset of V; a contradiction. Hence, rather there are only finitely many $n\geq 0$ with $I_{f_n^{-1}(V)}(x)=1$ , and hence $\lim _n f_n(x)=0$ , which is what we wanted to show.

Since $\lim _n (I_{f^{-1}_n(V)}\circ X)=I_{f^{-1}(V)}\circ X\, \mu $ -a.s., by Proposition 4.1(3) we have that $\lim _n I_{f^{-1}_n(V)}=I_{f^{-1}(V)}\, \mu _{X}$ -a.s. By the Dominated Convergence Theorem, $I_{f^{-1}_n(V)}\rightarrow I_{f^{-1}(V)}$ in $L_1(\mu _{X})$ .

We argue that $I_{f^{-1}_n(V)}$ is uniformly a computable point of $L_1(\mu _{X})$ . Since $f_n^{-1}(V)$ is c.e. open we have that $I_{f^{-1}_n(V)}$ is lsc. It further has computable $\mu _{X}$ expectation. For: by our hypothesis on the p-computable basis, we have that $f_n^{-1}(V)$ is both c.e. open and effectively closed, so that $\mathbb {E}_{\mu _{X}} I_{f_n^{-1}(V)} = \mu _{X}(f_n^{-1}(V))$ is both left-c.e. and right-c.e. and hence computable. Hence $I_{f^{-1}_n(V)}$ is an $L_1(\mu _{X})$ Schnorr test and so a computable point of $L_1(\mu _{X})$ .

Further, the $L_1(\mu _{X})$ distance between $I_{f^{-1}_n(V)}, I_{f^{-1}(V)}$ is uniformly computable:

(4.1) $$ \begin{align} & \|I_{f^{-1}_n(V)}- I_{f^{-1}(V)}\|_{L_1(\mu_{X})} =\mu_{X}(f^{-1}(V)\triangle f_n^{-1}(V)) =\mu\big(\Omega \times (f^{-1}(V)\triangle f_n^{-1}(V))\big)\notag \\& \quad = \mu\big((\Omega \times f^{-1}(V))\triangle (\Omega\times f_n^{-1}(V)\big) = \mu\big((V \times \mathcal{X})\triangle (\Omega\times f_n^{-1}(V))\big). \end{align} $$

To get the last identity, we use Proposition 4.1(1). Further, $\mu (V\times \mathcal {X})=p(V)$ , which is computable by hypothesis. And $\mu (\Omega \times f_n^{-1}(V))=\mu _{X}(f_n^{-1}(V))$ , which is computable by the previous paragraph. Hence, since (4.1) is the measure of a Boolean combination of c.e. open events in S with computable $\mu $ -measure, it too has $\mu $ -computable measure, and uniformly so.

Hence indeed $I_{f^{-1}(V)}$ is a computable point of $L_1(\mu _{X})$ , when $V\subseteq \Omega $ is an element of the p-computable basis.

Now suppose that U is a c.e. open with $p(U)$ computable. Then choose a computable sequence of elements $V_n$ from the p-computable basis such that $U=\bigcup _n V_n$ . By the previous argument, one has that $I_{f^{-1}(V_n)}$ is uniformly a computable point of $L_1(\mu _{X})$ . One has that:

(4.2) $$ \begin{align} & \|I_{f^{-1}(V_n)}- I_{f^{-1}(U)}\|_{L_1(\mu_{X})} =\mu_{X}(f^{-1}(V_n)\triangle f^{-1}(U)) =\mu\big(\Omega \times (f^{-1}(U)\triangle f^{-1}(V_n))\big) \notag \\& \quad = \mu\big((\Omega \times f^{-1}(U))\triangle (\Omega\times f^{-1}(V_n)\big) = \mu\big((U \times \mathcal{X})\triangle (V_n\times \mathcal{X})\big) = p(U\setminus V_n). \end{align} $$

The second to last identity follows again from Proposition 4.1(1). Hence the $L_1(\mu _{X})$ distance between $I_{f^{-1}(V_n)}, I_{f^{-1}(U)}$ is uniformly computable and goes to zero. Hence we can compute a subsequence such that $I_{f^{-1}(V_n)}\rightarrow I_{f^{-1}(U)}$ fast, so that $I_{f^{-1}(U)}$ is likewise a computable point of $L_1(\mu _{X})$ .

Proof. As a version of the conditional expectation, we use: $\mathbb {E}_{\mu _{X}}[f\mid \mathscr {F}_{n}](x):=\frac {1}{\mu _{X}([x_{n}])} \int _{[x_{n}]} f(x^{\prime }) \; d\mu _{X}(x^{\prime })$ when $\mu _{X}([x_{n}])>0$ , and zero otherwise.

Then $M_n(x)=\mu _{X}(B\mid x_{n})$ is a classical martingale since it is the conditional expectation of the random variable $I_B$ .

For the estimate, fix $m\geq 0$ . By Doob’s Maximal Inequality followed by Conditional Jensen one has

$$ \begin{align*} \|\sup_{m\leq n\leq k} \mu_{X}(B\mid \cdot_n)\|_{L_2(\mu_{X})} \leq \|\sup_{n\leq k} \mu_{X}(B\mid \cdot_n)\|_{L_2(\mu_{X})} \\ \leq 2\cdot \| \mu_{X}(B\mid \cdot_k)\|_{L_2(\mu_{X})}\leq 2\cdot \|I_{B}\|_{L_2(\mu_{X})}= 2\cdot \sqrt{\mu_{X}(B)}. \end{align*} $$

Then we are done by applying the Monotone Convergence Theorem on right-hand side to take $k\rightarrow \infty $ .

Proof. By Propositions 3.5 and 3.9 it suffices to show the part about pairs. Let $U\subseteq \Omega $ c.e. open with $p(U)$ computable.

We show $\sup _{n\geq m} p(U \hspace {-1mm}\mid \hspace {-1mm} x_{n})$ is a Schnorr $L_2(\mu _{X})$ -test, uniformly in $m\geq 0$ . By Proposition 3.7 and $0\leq p(U \hspace {-1mm}\mid \hspace {-1mm} x_{n})\leq 1$ , one has that $\sup _{n\geq m} p(U \hspace {-1mm}\mid \hspace {-1mm} x_{n})$ is lsc and $\leq 1$ , uniformly in $m\geq 0$ . Hence, it remains to show that $\sup _{n\geq m} p(U \hspace {-1mm}\mid \hspace {-1mm} x_{n})$ has computable $L_2(\mu _{X})$ -norm, uniformly in $m\geq 0$ . To this end, it suffices to show that given rational $\epsilon>0$ we can compute an index for a computable point of $L_2(\mu _{X})$ within $\epsilon $ of $\sup _{n\geq m} p(U \hspace {-1mm}\mid \hspace {-1mm} x_{n})$ .

Let $\epsilon>0$ be rational. By Proposition 4.2, compute an index for an element $V\subseteq U$ of the p-computable basis and compute $i\geq 0$ such that $\mu _{X}(f^{-1}(U)\triangle f^{-1}_i(V))<\frac {\epsilon ^2}{8}$ . Since $f_i^{-1}(V)$ is c.e. open, we can write $f_i^{-1}(V)=\bigsqcup _k [\sigma _k]$ , for a computable sequence of strings $\sigma _k$ in T. Since $f_i^{-1}(V)$ is also effectively closed, we have $\mu _{X}(f_i^{-1}(V))$ is computable. Hence, we can compute j such that $\mu _{X}(f_i^{-1}(V)\setminus \bigsqcup _{k\leq j} [\sigma _k])<\frac {\epsilon ^2}{8}$ . Putting these together, we have $\mu _{X}(f^{-1}(U)\triangle \bigsqcup _{k\leq j} [\sigma _k])<\frac {\epsilon ^2}{4}$ . Compute $\ell $ which is strictly greater than the maximum length of $\sigma _k$ for $k\leq j$ . Then one has the following, where the norm is with respect to $L_2(\mu _{X})$ and where the first inequality uses Proposition 4.1(2):

$$ \begin{align*} & \|\sup_{n\geq m} p(U \hspace{-1mm}\mid\hspace{-1mm} x_{n}) - \sup_{m\leq n \leq \ell} \mu_{X}(\bigsqcup_{k\leq j} [\sigma_k]\mid x_{n})\|_2 =\|\sup_{n\geq m} p(U \hspace{-1mm}\mid\hspace{-1mm} x_{n}) - \sup_{n\geq m} \mu_{X}(\bigsqcup_{k\leq j} [\sigma_k]\mid x_{n})\|_2 \\& \quad \leq \|\sup_{n\geq m} \mu_{X}(f^{-1}(U)\triangle \bigsqcup_{k\leq j} [\sigma_k]\mid x_{n})\|_2 \leq 2\cdot \sqrt{\mu_{X}(f^{-1}(U)\triangle \bigsqcup_{k\leq j} [\sigma_k])} <\epsilon. \end{align*} $$

The last two inequalities use Proposition 4.3 and our earlier estimate. Since the function $x\mapsto \sup _{m\leq n \leq \ell } \mu _{X}(\bigsqcup _{k\leq j} [\sigma _k]\mid x_{n})$ is a computable element of $L_2(\mu _{X})$ , we are done. Namely, we have shown that the function $x\mapsto \sup _{n\geq m} p(U \hspace {-1mm}\mid \hspace {-1mm} x_{n})$ is indeed a Schnorr $L_2(\mu _{X})$ -test, uniformly in $m\geq 0$ .

By change of variables, the function $(\theta ,x)\mapsto \sup _{n\geq m} p(U \hspace {-1mm}\mid \hspace {-1mm} x_{n})$ is a Schnorr $L_2(\mu )$ -test.

By the classical version of Lévy’s Upward Theorem, one has that $ \mu _{X}(f^{-1}(U)\mid x_{n})\rightarrow I_{f^{-1}(U)}(x)$ for $\mu _{X}$ -a.s. many x from $\mathcal {X}$ . Hence by Proposition 4.1(2) we have $p(U \hspace {-1mm}\mid \hspace {-1mm} x_{n})\rightarrow I_{U\times \mathcal {X}}(\theta , x)$ for $\mu $ -a.s. many $(\theta ,x)$ from S.

Then putting the two previous paragraphs together with Proposition 2.5, we have: for all $(\theta ,x)$ in $\mathsf {SR}^{\mu }\setminus (U\times \mathcal {X})$ that $p(U \hspace {-1mm}\mid \hspace {-1mm} x_{n}) \rightarrow 0$ .

Now it remains to show for all $(\theta ,x)$ in $\mathsf {SR}^{\mu }\cap (U\times \mathcal {X})$ that $p(U \hspace {-1mm}\mid \hspace {-1mm} x_{n}) \rightarrow 1$ .

Let $U=\bigcup _m U_m$ , where $U_m$ is a computable sequence of c.e. opens from the p-computable basis. Choose computable sequence $C_m\supseteq U_m$ of effectively closed sets such that $U_m, C_m$ have same p-measure.

It suffices to show that for each $m\geq 0$ and each $(\theta ,x)$ in $\mathsf {SR}^{\mu }\cap (U_m\times \mathcal {X})$ that $p(U_m \hspace {-1mm}\mid \hspace {-1mm} x_{n}) \rightarrow 1$ . For, given $(\theta ,x)$ in $\mathsf {KR}^{\mu }$ one has that $p(\cdot \hspace {-1mm}\mid \hspace {-1mm} x_{n})$ is in $\mathcal {P}(\Omega )$ by Propositions 3.4 and 3.6 and so this and $U_m\subseteq U$ also makes it that for each $m\geq 0$ and $(\theta ,x)$ in $\mathsf {SR}^{\mu }\cap (U_m\times \mathcal {X})$ one has $p(U \hspace {-1mm}\mid \hspace {-1mm} x_{n}) \rightarrow 1$ .

Fix $m\geq 0$ for the remainder of the proof. Let $V_m=\Omega \setminus C_m$ , which is c.e. open with computable p-measure. Then we can run the argument above to conclude that for all $(\theta ,x)$ in $\mathsf {SR}^{\mu }\setminus (V_m\times \mathcal {X})$ one has that $p(V_m \hspace {-1mm}\mid \hspace {-1mm} x_{n})\rightarrow 0$ . Let $(\theta ,x)$ in $\mathsf {SR}^{\mu }\cap (U_m\times \mathcal {X})$ . By Proposition 3.4, we have $\mathsf {SR}^{\mu }\subseteq \mathsf {KR}^{p}\times \mathcal {X}$ , and so $(\theta ,x)$ is in $\mathsf {SR}^{\mu }\setminus (V_m\times \mathcal {X})$ . Hence $p(V_m \hspace {-1mm}\mid \hspace {-1mm} x_{n})\rightarrow 0$ . Due to $(\theta ,x)$ in $\mathsf {KR}^{\mu }$ , one has $p(\cdot \mid x_{n})$ is in $\mathcal {P}(\Omega )$ by Propositions 3.4 and 3.6, so that $p(U_m \hspace {-1mm}\mid \hspace {-1mm} x_{n})=p(C_m \hspace {-1mm}\mid \hspace {-1mm} x_{n}) \rightarrow 1$ . Note the identity follows from the fact that for $(\theta ,x)$ in $\mathsf {KR}^{\mu }$ we have $p(\mathsf {KR}^{p} \hspace {-1mm}\mid \hspace {-1mm} x_{n})=1$ , as one can see from the formula for the posterior in (1.2).

Proof. Using the Effective Baire Category Theorem ([Reference Hoyrup and Rojas13, Theorem 5.1.2]), we can choose a p-computable basis consisting of intervals $(p,q)$ and $[0,p)$ and $(q,1]$ and $[0,1]$ such that $0\leq p<q\leq 1$ is a uniformly computable sequence of reals such that these $p,q$ are not equal to any element of the countable set ${\{\frac {1}{n}\sum _{i<n} \sigma (i) : \sigma \in 2^n, n>0\}}$ . This assumption makes both $I_{n,p,q} =\{\sigma \in 2^n: p<\frac {1}{n}\sum _{i<n} \sigma (i)<q\}$ and its relative complement $J_{n,p,q} =2^n\setminus I_{n,p,q}$ uniformly c.e. in $n,p,q$ . Further, one has $f_n^{-1}(p,q)=\bigsqcup _{\sigma \in I_{n,p,q}} [\sigma ]$ and $\mathcal {X} \setminus f_n^{-1}(p,q)=\bigsqcup _{\sigma \in J_{n,p,q}} [\sigma ]$ . Hence, $f_n^{-1}(p,q)$ is both c.e. open and effectively closed. The cases of $[0,p)$ and $(q,1]$ are similar. For $\epsilon>0$ and $n\geq 0$ one has:

$$ \begin{align*} & \mu(\left|f_n\circ X-\Theta\right|>\epsilon) = \mu(\left|\sum_{i<n} (x(i)-\theta)\right|>n\cdot \epsilon) = \mu(\left|\sum_{i<n} (x(i)-\theta)\right|^2>n^2\cdot \epsilon^2)\\ & \leq \frac{1}{\epsilon^2\cdot n^2} \mathbb{E}_{\mu}\big(\sum_{i<n} (g_i-\Theta)\big)^2 = \frac{1}{\epsilon^2\cdot n^2} \sum_{i<n} \mathbb{E}_{\mu} (g_i-\Theta)^2 = \frac{1}{\epsilon^2\cdot n} (\alpha-\beta)\\[-49pt] \end{align*} $$

Proof. Since $f_n, \Theta $ are computable continuous, so is $\left |f_n\circ X -\Theta \right |$ . By Proposition 2.8, choose a computable sequence of reals $\eta _n$ for $n\geq 1$ such that $\eta _n$ is in the interval $((n+1)^{-\frac {1}{3}}, n^{-\frac {1}{3}})$ and such that the event $\left |f_{n^2}\circ X -\Theta \right |>\eta _n$ has computable $\mu $ -measure. Then let $U_n$ be this event, and set $g=\sum _{n\geq 1} I_{U_n}$ . Then one has $\mu (U_n)\leq \mu \big (\left |f_{n^2}\circ X -\Theta \right |>(n+1)^{-\frac {1}{3}}\big )\leq (n+1)^{\frac {2}{3}}\cdot n^{-2} \cdot (\alpha -\beta ) \leq \sqrt [3]{4} \cdot n^{-\frac {4}{3}} \cdot (\alpha -\beta )$ . Then one has $\mathbb {E}_{\mu } g \leq \sum _{n=1}^{\infty } \sqrt [3]{4} \cdot n^{-\frac {4}{3}} \cdot (\alpha -\beta )$ , and since the latter is finite and computable so is $\mathbb {E}_{\mu } g$ by the Comparison Test. Hence g is an $L_1(\mu )$ Schnorr test.

Suppose that $(\theta ,x)$ is in $\mathsf {SR}^{\mu }$ . Choose $n_0> 0$ such that for all $n\geq n_0$ one has $(\theta ,x)$ is not in $U_n$ . Then for all $n\geq n_0$ we have $\left |\sum _{i<n^2} (x(i)-\theta )\right |\leq n^{\frac {5}{3}}$ .

Let $m> n_0^2$ . Choose least $n> n_0$ with $n^2<m\leq (n+1)^2$ . Then one has

$$ \begin{align*} \left|\sum_{i<m} (x(i)-\theta)\right| \leq \left|\sum_{i<n^2} (x(i)-\theta)\right|+(n+1)^2-n^2 \leq n^{\frac{5}{3}}+2n+1 \leq m^{\frac{5}{6}}+2\sqrt{m}+1 \end{align*} $$

Hence $\left |(\frac {1}{m} \sum _{i<m} x(i))-\theta \right |=\left |\frac {1}{m} \sum _{i<m} (x(i)-\theta )\right |\leq m^{-\frac {1}{6}}+2\cdot m^{-\frac {1}{2}}+m^{-1}$ .

Proof. The direction from (1) to (2) follows directly from Theorem 1.9.

We now show that “not (1)” implies “not (2).” Suppose that $\theta _0$ is in $\bigcap _n U_n$ , where $U_n$ is a sequential Schnorr p-test. Without loss of generality $U_0=\Omega $ and $U_n$ is decreasing. Note that since $U_0=\Omega $ , we have $\sum _n p(U_n\setminus U_{n+1})=1$ and hence $\sum _n \mu ((U_n\setminus U_{n+1})\times \mathcal {X})=1$ .

By the hypothesis of the Theorem, choose an apt triple $\mathcal {X}, \theta \mapsto p(\cdot \hspace {-1mm}\mid \hspace {-1mm} {\theta }), f_n:\mathcal {X}\rightarrow \Omega $ (cf. Definition 1.12). We will construct another apt triple $\mathcal {Y}, \theta \mapsto q(\cdot \hspace {-1mm}\mid \hspace {-1mm} {\theta }), g_n:\mathcal {Y}\rightarrow \Omega $ such that $\theta _0$ is not computably consistent relative to $p, \mathcal {Y}, \theta \mapsto q(\cdot \hspace {-1mm}\mid \hspace {-1mm} {\theta })$ .

Let $\mathcal {Y}=[T^{\infty }]$ , where $T^{\infty }=\{\emptyset \}\cup \{(m)^{\frown }\sigma : m\geq 0, \sigma \in T\}$ , where $\emptyset $ denotes the root (the unique length zero string). That is, $T^{\infty }$ copies T countably many times after the root $\emptyset $ . Clearly since T is a computable subtree of $\mathbb {N}^{<\mathbb {N}}$ with no dead ends, so is $T^{\infty }$ .

Define likelihood as follows, for $\tau $ in $T^{\infty }$ :

(6.1) $$ \begin{align} q(\tau \hspace{-1mm}\mid\hspace{-1mm} \theta) = \begin{cases} 1 & \text{if } \tau \text{ is the root and } \theta \text{ is not in } \bigcap_n U_n, \\ 0 & \text{if } \tau \text{ is the root and } \theta \text{ is in } \bigcap_n U_n, \\ p(\sigma \hspace{-1mm}\mid\hspace{-1mm} \theta)\cdot I_{U_m\setminus U_{m+1}}(\theta) & \text{if } \tau=(m)^{\frown}\sigma. \end{cases} \end{align} $$

The idea is: if $\theta $ is in $U_m\setminus U_{m+1}$ , then $q(\cdot \hspace {-1mm}\mid \hspace {-1mm} {\theta })$ puts all the probability on the m-th copy of T in $T^{\infty }$ , where it copies the original likelihood; whereas if $\theta $ is in the p-null set $\bigcap _n U_n$ , then $q(\cdot \hspace {-1mm}\mid \hspace {-1mm} {\theta })$ is the zero element of $\mathcal {M}^+(\mathcal {Y})$ .

Regarding the effective properties of the likelihood function (cf. Definition 1.1(3), note that for p-a.s. many $\theta $ in $\Omega $ , one has that $\theta $ is not in $\bigcap _n U_n$ and hence is in some $U_m\setminus U_{m+1}$ , and thus:

• – The likelihood function $\theta \mapsto q(\tau \hspace {-1mm}\mid \hspace {-1mm} \theta )$ is $L_1(p)$ computable, uniformly in $\tau $ in $T^{\infty }$ by Proposition 2.6(3) and Proposition 2.7.

• – $q(\cdot \hspace {-1mm}\mid \hspace {-1mm} {\theta })$ is in $\mathcal {P}(\mathcal {Y})$ for p-a.s. many $\theta $ in $\Omega $ since $p(\cdot \hspace {-1mm}\mid \hspace {-1mm} {\theta })$ is in $\mathcal {P}(\mathcal {X})$ for p-a.s. many $\theta $ in $\Omega $ .

Suppose that, per Definition 1.6, $p(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )$ is understood pointwise as its version $g(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )-h(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )$ , where these are $L_1(p)$ Schnorr tests, uniformly in $\sigma $ from T. We define a pair $G(\tau \hspace {-1mm}\mid \hspace {-1mm} \theta ), H(\tau \hspace {-1mm}\mid \hspace {-1mm} \theta )$ of $L_1(p)$ Schnorr tests, uniformly in $\tau $ in $T^{(\infty )}$ , such that $G(\tau \hspace {-1mm}\mid \hspace {-1mm} \theta )-H(\tau \hspace {-1mm}\mid \hspace {-1mm} \theta )$ is a version of $q(\tau \hspace {-1mm}\mid \hspace {-1mm} \theta )$ :

• – If $\tau $ is the root, define $G(\tau \hspace {-1mm}\mid \hspace {-1mm} \theta ) = \sum _m I_{U_{m}}(\theta )$ and $H(\tau \hspace {-1mm}\mid \hspace {-1mm} \theta ) = \sum _m I_{U_{m+1}}(\theta )$ .

• – If $\tau = (m)^{\frown }\sigma $ , define $G((m)^{\frown }\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta ) =g(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )\cdot I_{U_m}(\theta )+h(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )\cdot I_{U_{m+1}}(\theta )$ and $H((m)^{\frown }\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta ) =g(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )\cdot I_{U_{m+1}}(\theta )+h(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )\cdot I_{U_m}(\theta )$ .

Then $G(\tau \hspace {-1mm}\mid \hspace {-1mm} \theta ), H(\tau \hspace {-1mm}\mid \hspace {-1mm} \theta )$ are finite for $\theta $ in $\mathsf {SR}^{p}$ , and on these points we have that $G(\tau \hspace {-1mm}\mid \hspace {-1mm} \theta )-H(\tau \hspace {-1mm}\mid \hspace {-1mm} \theta )$ agrees with (6.1), when $p(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )$ is understood pointwise as its version $g(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )-h(\sigma \hspace {-1mm}\mid \hspace {-1mm} \theta )$ . When $\theta $ is in $\bigcap _n U_n$ , by our convention that $\infty -\infty =0$ , we have that $G(\tau \hspace {-1mm}\mid \hspace {-1mm} \theta )-H(\tau \hspace {-1mm}\mid \hspace {-1mm} \theta )=0$ for all $\tau $ . Hence when $\theta $ is in $\bigcap _n U_n$ , this version of $q(\cdot \hspace {-1mm}\mid \hspace {-1mm} {\theta })$ is the zero element of $\mathcal {M}^+(\mathcal {Y})$ , and hence not an element of $\mathcal {P}(\mathcal {Y})$ .Footnote ¹⁸ By the first conjunct of Definition 1.8(2), any $\theta $ in $\bigcap _n U_n$ is not computably consistent.

Define uniformly computable continuous $g_n:\mathcal {Y}\rightarrow \Omega $ by $g_n((m)^{\frown } x) = f_n(x)$ . It remains to show that $\mathcal {Y}, \theta \mapsto q(\cdot \hspace {-1mm}\mid \hspace {-1mm} {\theta }), g_n$ satisfies conditions (1)-(2) of Theorem 1.9.

To this end, define uniformly computable continuous $\iota _m:S\rightarrow \Omega \times \mathcal {Y}$ by $\iota _m(\theta , x)=(\theta , (m)^{\frown } x)$ . Further, let $\nu $ be the joint distribution on the product space $\Omega \times \mathcal {Y}$ formed from the prior p and the new likelihood $\theta \mapsto q(\tau \hspace {-1mm}\mid \hspace {-1mm} \theta )$ . Then for each Borel event $B\subseteq \Omega \times \mathcal {Y}$ , we have

(6.2) $$ \begin{align} \nu(B) = \sum_m \mu(\iota_m^{-1}(B)\cap \big((U_m\setminus U_{m+1})\times \mathcal{X}\big)) \end{align} $$

This follows by a simple induction on B, which we omit for reasons of space.

Finally, here is the verification that $\mathcal {Y}, \theta \mapsto q(\cdot \hspace {-1mm}\mid \hspace {-1mm} {\theta }), g_n$ satisfies conditions (1)-(2) of Theorem 1.9:

1. – Suppose that A is a $\mu $ -measure one subset of $\mathsf {KR}^{p}\times \mathcal {X}$ on which $\lim _n f_n(x)$ exists and is equal to $\theta $ . Let $B=\bigsqcup _m \iota _m(A)$ , so that $\iota _m^{-1}(B)=A$ for all $m\geq 0$ and so that $B\subseteq \mathsf {KR}^{p}\times \mathcal {Y}$ . Then by (6.2) one has

   $$ \begin{align*} \nu(B) = \sum_m \mu(A\cap \big((U_m\setminus U_{m+1})\times \mathcal{X}\big)) = \sum_m \mu((U_m\setminus U_{m+1})\times \mathcal{X}) =1 \end{align*} $$
   Finally note that for all $(\theta , (m)^{\frown } x)$ in B one has that $(\theta , x)$ is in A, and hence by condition (1) of Theorem 1.9, we have that the limit $\lim _n g_n((m)^{\frown } x))=\lim _n f_n(x)$ exists and is equal to $\theta $ .

2. – Suppose that $U\subseteq \Omega $ is from a p-computable basis. Then, by condition (2) of Theorem 1.9 we can write $f_n^{-1}(U)=\bigsqcup _i [\sigma _i]$ and $\mathcal {X}\setminus f_n^{-1}(U)=\bigsqcup _i [\tau _i]$ for computable sequences of $\tau _i, \sigma _i$ from T. Then one has $g_n^{-1}(U)=\bigsqcup _m \bigsqcup _i [(m)^{\frown } \sigma _i]$ and $\mathcal {Y}\setminus g_n^{-1}(U)=\bigsqcup _m \bigsqcup _i [(m)^{\frown } \tau _i]$ .

Proof. We identify the countable dense set D of Y with natural numbers and use variables $i,j$ for them in this proof.

Since classically $d_0$ provides a complete compatible metric for $Y_0$ (cf. [Reference Kechris16, p. 17]), to show that $Y_0$ is a computable Polish space it remains to show that if $i,j$ come from $D_0$ , then $d_0(i,j)$ is uniformly computable. To see this, note that since $i,j$ are computable points of $Y_0\subseteq U_n$ , we have that $d(i,C_n), d(j,C_n)>0$ are uniformly computable. We can search for, and find, a non-zero rational, uniformly in $i,j,n$ , which is below $d(i,C_n), d(j,C_n)$ . Using this, we can compute $\min \big (2^{-(n+1)}, \left |\frac {1}{d(i,C_n)}-\frac {1}{d(j,C_n)}\right |\big )$ . Then by the Comparison Test, $d_0(i,j)$ is computable, and uniformly so. Hence indeed $Y_0$ is a computable Polish space.

Now we show that $Y_0$ is a computable Polish subspace of Y.

We focus on condition (3) of Definition 7.1 since the other one is similar but easier. It suffices to consider the case where V is an open ball $B_{d_0}(j,\epsilon ):=\{y_0\in Y_0: d_0(j,y_0)<\epsilon \}$ in $Y_0$ , where j in $D_0$ and $\epsilon>0$ is rational; we must show that $B_{d_0}(j, \epsilon )=U\cap Y_0$ for some c.e. open U in Y. We proceed as follows:

1. – For each rational q in $(0,\frac {1}{c})$ compute $n_q\geq 0$ such that $2^{-(n_q+1)}<\frac {q}{3}$ .

2. – For each $(i,q)$ in $D_0\times (\mathbb {Q}\cap (0,\frac {1}{c}))$ , compute real numbers

   $$ \begin{align*} \delta_{i,q,n}:=\frac{1}{d(i,C_n)}+ \frac{q}{3}\cdot 2^{-(n+1)}, \hspace{10mm} \eta_{i,q,n}:=\frac{1}{d(i,C_n)}- \frac{q}{3}\cdot 2^{-(n+1)} \end{align*} $$
   Since the metric d is bounded $<c$ and since q is from $(0,\frac {1}{c})$ , note that we have that $\eta _{i,q,n}>0$ ; and trivially $\delta _{i,q,n}>0$ .

3. – Define the c.e. index set $I=\{(i,q)\in D_0\times (\mathbb {Q}\cap (0,\frac {1}{c})): d_0(i,j)+\frac {q}{3}<\epsilon \}$ .

4. – For each $(i,q)$ in I let $U_{i,q,n}$ be the c.e. open

   (7.2) $$ \begin{align} U_{i,q,n}=(d(\cdot, C_n))^{-1}(-\infty,\frac{1}{\eta_{i,q,n}})\cap (d(\cdot, C_n))^{-1}(\frac{1}{\delta_{i,q,n}},\infty) \end{align} $$

Then we claim that $B_{d_0}(j,\epsilon )=Y_0\cap \bigcup _{(i,q)\in I} B_d(i,\frac {q}{3})\cap \bigcap _{n\leq n(q)} U_{i,q,n}$ :

• • First suppose that x in $Y_0\cap B_d(i,\frac {q}{3})\cap \bigcap _{n\leq n(q)} U_{i,q,n}$ for some $(i,q)$ in I. Then for $n\leq n(q)$ , membership in $U_{i,q,n}$ implies that $\left |\frac {1}{d(i,C_n)}-\frac {1}{d(x,C_n)}\right |<\frac {q}{3}\cdot 2^{-(n+1)}$ . Together with membership in $ B_d(i,\frac {q}{3})$ and the definition of $n_q$ we then have that $d_0(x,i)<q$ . Then $d_0(x,j)\leq d_0(x,i)+d_0(i,j)<q+d_0(i,j)<\epsilon $ .

• • Second suppose that x is in $B_{d_0}(j,\epsilon )$ . Choose non-zero rational q such that both $q<\frac {3}{2}(\epsilon -d_0(x,j))$ and $q<\frac {1}{c}$ . Since for each $n\leq n_q$ , the function $y\mapsto \frac {1}{d(y,C_n)}$ is continuous on $U_n$ and x is in $U_n$ , we can find i in $D_0$ such that x is in $B_{d_0}(i,\frac {q}{3})$ and x is in $U_{i,q,n}$ for all $n\leq n_q$ . Further $(i,q)$ is in I: for we have $d_0(i,j)+\frac {q}{3}\leq d_0(i,x)+d_0(x,j)+\frac {q}{3}<\frac {2}{3}q+d_0(x,j)<\epsilon $ .