Two-stage conditional-error adaptation

Fix $u\in [0,1]$ and $H\in {\mathcal{F}}_1$. The event $\{p_N\le u\}$ is measurable by the p-value statistic definition [F2]. The events $D_m$, $m\in M$, are pairwise disjoint and partition $\mathrm{\Omega }$. Countable additivity [F3] gives $P_0(H\cap \{p_N\le u\})={\sum }_{m\in M}{P}_0(H\cap D_m\cap \{p_N\le u\})$ and $P_0(H)={\sum }_{m\in M}{P}_0(H\cap D_m)$. Summing the assumed stratum inequalities over the increasing enumeration of $M$ (equivalently, taking finite partial sums and then their limits) yields $P_0(H\cap \{p_N\le u\})\le u{P}_0(H)$.

Let $Y_u$ be a version of $E_0[{\mathbf{1}}_{\{p_N\le u\}}\mid {\mathcal{F}}_1]$, which exists by [A1] and [F1]. The event identities in [F1] and [F4] turn step 1.1 into ${\int }_H{Y}_{u}d{P}_0\le u{P}_0(H)$ for every $H\in {\mathcal{F}}_1$. Also $Y_u\ge 0$ almost surely by [F5]. If $P_0(Y_u>u)>0$, then for some integer $k\ge 1$ the set $H_k=\{Y_u\ge u+1\mathrm{/}k\}\in {\mathcal{F}}_1$ has positive probability, because $\{Y_u>u\}={\bigcup }_{k\ge 1}{H}_k$. Monotonicity [F6] gives ${\int }_{H_k}{Y}_{u}d{P}_0\ge (u+1\mathrm{/}k)P_0(H_k)>u{P}_0(H_k)$, contradicting the preceding event-integral bound. Thus $Y_u\le u$ almost surely. The argument is for each fixed $u$ and does not assert a common exceptional set over uncountably many thresholds.

By [A1] and [F1], both conditional-expectation classes exist. Since $0\le {\mathbf{1}}_{R_j}\le 1$ for $j=0,1$, order preservation and the constant rule in [F2] give $0\le A,Q\le 1$ almost surely. All representatives of either class agree almost surely by [F1], so the rejection probabilities do not depend on the chosen versions.

Let ${\mathcal{F}}_0=\{\mathrm{\varnothing },\mathrm{\Omega }\}$. By [F3], $E_0[A\mid {\mathcal{F}}_0]=E_0[{\mathbf{1}}_{R_0}\mid {\mathcal{F}}_0]$. Each side is constant almost surely because it is ${\mathcal{F}}_0$-measurable; testing its defining event identity at $\mathrm{\Omega }$ shows its constant is its ordinary expectation. Thus [F4] gives $E_{0}A=E_0{\mathbf{1}}_{R_0}=P_0(R_0)\le \alpha$. The same argument gives $E_{0}Q=P_0(R_1)$. Order preservation [F2], applied to $Q\le A$ with conditioning sigma-algebra ${\mathcal{F}}_0$, gives $E_0[Q\mid {\mathcal{F}}_0]\le E_0[A\mid {\mathcal{F}}_0]$; these constants are $E_{0}Q$ and $E_{0}A$, so $P_0(R_1)=E_{0}Q\le E_{0}A\le \alpha$.

Conditional order and the constant rule [F5] give $0\le A\le 1$ almost surely. Clipping a version on the measurable null set where it leaves $[0,1]$ preserves its conditional-expectation class and changes $R_2$ only on a null set, so take $0\le A\le 1$ everywhere. For each integer $n\ge 1$, define $A_n=2^{-n}\lceil 2^{n}A\rceil$. Its level sets are $\{A_n=0\}=\{A=0\}$ and $\{A_n=k{2}^{-n}\}=\{(k-1)2^{-n}<A\le k{2}^{-n}\}$ for $1\le k\le 2^n$, so they are in ${\mathcal{F}}_1$. The inequality $\lceil 2x\rceil \le 2\lceil x\rceil$ shows $A_{n+1}\le A_n$, and $0\le A_n-A<2^{-n}$, hence $A_n$ is finite-valued and $A_n\downarrow A$. Write its values as $t_{n,k}=k{2}^{-n}$, $k=0,\dots ,2^n$, and put $S_{n,k}=\{A_n=t_{n,k}\}$. By [F2], the events $\{p_2\le t_{n,k}\}$ are measurable. Let $Q_n=E_0[{\mathbf{1}}_{\{p_2\le A_n\}}\mid {\mathcal{F}}_1]$. The disjoint partition by $S_{n,k}$, linearity [F5], and taking out the bounded known indicator [F3] give $Q_n={\sum }_{k=0}^{2^n}{\mathbf{1}}_{S_{n,k}}{E}_0[{\mathbf{1}}_{\{p_2\le t_{n,k}\}}\mid {\mathcal{F}}_1]\le {\sum }_{k=0}^{2^n}{t}_{n,k}{\mathbf{1}}_{S_{n,k}}=A_n$ almost surely. For all $n$ these inequalities hold on one probability-one set, because only countably many fixed dyadic thresholds occur.

The events $B_n=\{p_2\le A_n\}$ decrease to $B=\{p_2\le A\}$. Their complement indicators are bounded and hence integrable, and increase to ${\mathbf{1}}_{B^c}$. Thus [F4] gives $E_0[{\mathbf{1}}_{B_n^c}\mid {\mathcal{F}}_1]\uparrow E_0[{\mathbf{1}}_{B^c}\mid {\mathcal{F}}_1]$ almost surely. By linearity and the constant rule [F5], this is equivalent to $Q_n\downarrow Q:=E_0[{\mathbf{1}}_B\mid {\mathcal{F}}_1]$. Taking limits in step 1.1 and using $A_n\downarrow A$ yields $Q\le A$ almost surely.

Apply [F6] to $R_2=B$: its conditional rejection probability is the $Q$ of step 2.1 and is at most $A$, while the original conditional-error budget has expectation at most $\alpha$. Therefore $P_0(R_2)\le \alpha$.