BY arbitrary dependence and BH PRDS

The interval is closed in $\mathbb{R}$: for $z\mathrm{\notin }I$, [F6] and [A1] give either $z<0$ or $z>1$. If $z<0$, set $r=-z>0$; then every $y\in B(z,r)$ satisfies $y<z+r=0$. If $z>1$, set $r=z-1>0$; then every $y\in B(z,r)$ satisfies $y>z-r=1$. By [F1] these are open neighborhoods disjoint from $I$, so its complement is open by [F10]. Thus $I$ is Borel by [F7].

The set $D:=\mathbb{Q}\cap I$ is at most countable: restrict the bijection $\mathbb{Q}\to \mathbb{N}$ from [F3] to $D$, then apply [F5]. It is dense in $I$. Indeed, for $x\in I$ and $\epsilon >0$, if $x=1$ choose $q=1$. If $x<1$, then $x+\epsilon >x$; take $b=x+\epsilon$ when $x+\epsilon \le 1$, and $b=1$ otherwise. In either case $b>x$ and $b-x\le \epsilon$, with strict inequality in the second case. By [F4] choose a rational $q$ with $x<q<b$; then $q\in D$ and $\mathrm{\mid }q-x\mathrm{\mid }<\epsilon$. The order comparisons use [F6] and the interval endpoints use [A1].

Let $(x_n)$ be a Cauchy sequence in $I$ under $d_I$. By [A2]–[A3], Cauchy distances are the same in $I$ and $\mathbb{R}$, so [F2] gives a real limit $x$. If $x\mathrm{\notin }I$, step 1.1 gives an open ball about $x$ disjoint from $I$, while convergence puts all sufficiently late $x_n$ in that ball, a contradiction. Hence $x\in I$ and the sequence converges in the restricted metric; $(I,d_I)$ is complete.

By steps 1.2 and 2.1, $I$ is separable and completely metrizable, hence Polish by [F9]. Its Borel sigma-algebra is the trace of $\mathcal{B}(\mathbb{R})$ by [F8] and [A2]. The identity map from $(I,\mathcal{B}(\mathbb{R}){\mathrm{\mid }}_I)$ to the Polish subspace $I$ is therefore a bimeasurable bijection; [F9] makes $(I,\mathcal{B}(I))$ standard Borel. This proves the claim without any choice axiom.

For $1\le r\le m-1$ and $t\in \mathbb{R}$, $\{y_{(r)}\le t\}$ is the finite union over $r$-element index sets $J$ of ${\bigcap }_{j\in J}\{y_j\le t\}$, since at least $r$ remaining values are at most $t$ exactly when the $r$th ordered value is at most $t$. Each coordinate cylinder is measurable in the stated finite product structure by [F1] and [F3], so these sublevel sets are Borel. Since the closed left rays generate $\mathcal{B}(\mathbb{R})$ by [F4], each order statistic is Borel by [F2]. The coordinate-deletion projection from $[0,1{]}^m$ to $[0,1{]}^{m-1}$ is measurable: preimages of measurable rectangles are finite intersections of coordinate cylinders, and the rectangles generate the product sigma-algebra by [F3]. Thus lifting a Borel set in the remaining-coordinate space gives a Borel set in the full product. If $y\le z$ coordinatewise, at least $r$ coordinates of $z$ are at most $z_{(r)}$; their corresponding coordinates of $y$ are also at most $z_{(r)}$, so $y_{(r)}\le z_{(r)}$. Every remaining-coordinate order statistic is therefore coordinatewise nondecreasing.

Fix $k$ and suppose $p_i\le c_k$. Rank $k$ passes exactly when at least $k-1$ other coordinates are at most $c_k$, which is $y_{(k-1)}\le c_k$; for $k=1$ this holds by $y_{(0)}=0$. For each $j>k$, since $p_i\le c_k\le c_j$, rank $j$ passes exactly when at least $j-1$ other coordinates are at most $c_j$, namely $y_{(j-1)}\le c_j$. Hence $R=k$ exactly when the first condition of $C_{i,k}$ holds and every higher rank fails, which is $p_{-i}\in C_{i,k}$. If $k=m$, $i$ is automatically among the first $k$ ranks. If $k<m$ and $i$ is not among the first $k$ ranks, then $p_{(k+1)}\le p_i\le c_k\le c_{k+1}$ would make rank $k+1$ pass. Conversely, if $R=k$ and $i$ is rejected, then $p_i\le p_{(k)}\le c_k$, rank $k$ passes and every higher rank fails, so the conditions defining $C_{i,k}$ hold.

For any remaining vector $y$, replace coordinate $i$ by $0$. Since $c_1>0$, rank one passes. Every rank occupied by a zero-valued coordinate also passes because all critical values are positive, so $R$ includes all zero-valued coordinates and rejects the inserted coordinate $i$. Thus the resulting rejection count is some $r\in \{1,\dots ,m\}$. By step 2.1, $y\in C_{i,r}$; hence the sets cover the space. If $k<\mathrm{\ell }$ and $y\in C_{i,k}$, its condition at $r=\mathrm{\ell }-1$ gives $y_{(\mathrm{\ell }-1)}>c_{\mathrm{\ell }}$, while membership in $C_{i,\mathrm{\ell }}$ would give $y_{(\mathrm{\ell }-1)}\le c_{\mathrm{\ell }}$. Thus the sets are disjoint. They are Borel by step 1.1 and their finite defining inequalities.

If $y\in C_{i,j}$ for some $j\le k<m$, the defining inequalities include $y_{(r)}>c_{r+1}$ for every $r\ge k$, so $D_{i,k}$ is contained in the displayed intersection. Conversely, suppose those inequalities hold and set $p_i=0$. Rank one passes, and by the zero-coordinate argument in step 3.1 the inserted coordinate $i$ is rejected. No rank $\mathrm{\ell }\ge k+1$ can pass: that would require $y_{(\mathrm{\ell }-1)}\le c_{\mathrm{\ell }}$, contrary to the assumed condition at $r=\mathrm{\ell }-1$. Thus the rejection count is some $j\le k$, and step 2.1 puts $y$ in $C_{i,j}$. This proves the formula; when $k=m$, step 3.1 gives $D_{i,m}$ equal to the whole space. The formula is a finite intersection of Borel order-statistic superlevel sets and is coordinatewise increasing by step 1.1.

Fix $\theta \in \mathrm{\Theta }$ and let $I_i$ indicate that BY rejects hypothesis $i$. Pointwise, $V_{\theta }\mathrm{/}(R\vee 1)={\sum }_{i\in I_0(\theta )}{\sum }_{k=1}^m{k}^{-1}\mathbf{1}\{I_i=1,R=k\}$: if $R=0$ both sides are zero, and if $R>0$ only $k=R$ contributes, with weight $1\mathrm{/}R$ for each rejected true null. This finite sum is bounded and integrable by [F7]; [F5] and [F6] therefore give ${\mathrm{FDR}}_{\theta }={\sum }_{i\in I_0(\theta )}{\sum }_{k=1}^m{k}^{-1}{P}_{\theta }(I_i=1,R=k)$.

Fix $i\in I_0(\theta )$ and put $t=q\mathrm{/}(m{H}_m)$ and $J_j=((j-1)t,jt]$ for $j=1,\dots ,m$. Since $\theta \in H_{0i}$, that null is nonempty and [F4] at $u=0$ gives $P_{\theta }(p_i=0)=0$. Since BY has $c_k=kt$, [F3] splits each event $\{I_i=1,R=k\}$ into the disjoint events $\{p_i\in J_j,p_{-i}\in C_{i,k}\}$ for $j\le k$, apart from that null zero event. Using $1\mathrm{/}k\le 1\mathrm{/}j$ for $k\ge j$ and that the $C_{i,k}$ partition the remaining coordinates, ${\sum }_{k=1}^m{k}^{-1}{P}_{\theta }(I_i=1,R=k)={\sum }_{j=1}^m{\sum }_{k=j}^m{k}^{-1}{P}_{\theta }(p_i\in J_j,p_{-i}\in C_{i,k})\le {\sum }_{j=1}^m{j}^{-1}{P}_{\theta }(p_i\in J_j)$.

Set $F_j=P_{\theta }(p_i\le jt)$ for $j=0,\dots ,m$. Then $F_0=P_{\theta }(p_i=0)=0$, $F_j\le jt$ by [F4], and $P_{\theta }(p_i\in J_j)=F_j-F_{j-1}$. Finite summation by parts gives ${\sum }_{j=1}^m(F_j-F_{j-1})\mathrm{/}j=F_m\mathrm{/}m+{\sum }_{j=1}^{m-1}{F}_j(1\mathrm{/}j-1\mathrm{/}(j+1))\le t+t{\sum }_{j=1}^{m-1}1\mathrm{/}(j+1)=t{H}_m=q\mathrm{/}m$. Here $mt=q\mathrm{/}{H}_m\le q<1$, and $H_m={\sum }_{j=1}^{m}1\mathrm{/}j$.

Combining steps 1.2 and 2.1 bounds each true-null contribution in step 1.1 by $q\mathrm{/}m$. Summing over the $m_0(\theta )$ true nulls gives ${\mathrm{FDR}}_{\theta }\le (m_0(\theta )\mathrm{/}m)q\le q$; with no true nulls the outer sum is empty and FDR is zero. No dependence condition is used, and the CDF summation-by-parts argument covers discrete superuniform p-values.

Put $c_k=kq\mathrm{/}m$, $A_k:=\{P_i\le c_k\}$, and $B_{i,k}:=\{P_{-i}\in C_{i,k}\}$. If $P(A_k)>0$, set $a_k:=P(A_k\cap B_{i,k})\mathrm{/}P(A_k)$; otherwise set $a_k=0$. By [F2], $I_i\mathrm{/}(R\vee 1)={\sum }_{k=1}^m{k}^{-1}{\mathbf{1}}_{A_k\cap B_{i,k}}$. The terms are bounded and the sum is finite, so [F5]–[F7] give $\mathbb{E}[I_i\mathrm{/}(R\vee 1)]={\sum }_{k=1}^m(P(A_k)\mathrm{/}k)a_k$; a zero-probability $A_k$ contributes zero by the definition of $a_k$.

For each $k$, lift $D_{i,k}$ to ${\tilde{D}}_{i,k}:=\{p\in [0,1{]}^m:p_{-i}\in D_{i,k}\}$. The coordinate-deletion map from $[0,1{]}^m$ to the remaining-coordinate product is measurable: the preimage of each measurable rectangle is a finite intersection of coordinate cylinders, and such rectangles generate the product sigma-algebra. Hence ${\tilde{D}}_{i,k}$ is Borel, and it is coordinatewise increasing. Let $g_k(x):=K_i(x,{\tilde{D}}_{i,k})$, nondecreasing on the support by [F1] and [A1]; the kernel exists on this standard-Borel interface under [A2].

Write $S_i=\mathrm{supp}({\mu }_i)$. The family $\mathcal{U}:=\{(a,b)\cap [0,1]:a,b\in \mathbb{Q},a<b\}$ is a basis for the relative topology on $[0,1]$. If $U$ is relative open and $x\in U$, write $U=O\cap [0,1]$ with $O$ open in $\mathbb{R}$ by [F8]; choose $\epsilon >0$ with $(x-\epsilon ,x+\epsilon )\subseteq O$ using [F8], then use rational density from [F8] to choose $a,b\in \mathbb{Q}$ with $x-\epsilon <a<x<b<x+\epsilon$. Thus $x\in (a,b)\cap [0,1]\subseteq U$. Since $\mathbb{Q}\times \mathbb{Q}$ is nonempty and at most countable by [F8], [F9] gives an enumeration $((a_n,b_n){)}_{n\in \mathbb{N}}$ of it, whose interval images enumerate $\mathcal{U}$; hence the basis is countable. Let $E_n=(a_n,b_n)\cap [0,1]$ if $a_n<b_n$ and ${\mu }_i((a_n,b_n)\cap [0,1])=0$, and $E_n=\mathrm{\varnothing }$ otherwise. By the definition of support in Positive regression dependence on true-null coordinates (PRDS), $[0,1]\setminus S_i={\bigcup }_{n\in \mathbb{N}}{E}_n\mathrm{.}$ Indeed, a point outside the support has a relative open null neighborhood and hence a basis neighborhood contained in it, which occurs in the enumeration; conversely, membership in any such null basis neighborhood excludes support membership. The right-hand side is a countable union of null Borel sets, so [F8] gives ${\mu }_i([0,1]\setminus S_i)=0$. Thus $S_i$ is Borel and ${\mu }_i(S_i)=1$.

By [F4], whenever $P(A_k)>0$, $\frac{P(A_k\cap \{P_{-i}\in D_{i,k}\})}{P(A_k)}=\frac{{\int }_{[0,c_k]}{g}_k(x){\mu }_i(dx)}{{\mu }_i([0,c_k])}\mathrm{.}$ For fixed $k<m$, this average is at most the corresponding average over $[0,c_{k+1}]$. By step 1.3, both averages may be restricted to $S_i$ without changing them. Every supported $x\le c_k$ is below every supported $y\in (c_k,c_{k+1}]$, so $g_k(x)\le g_k(y)$ by [F1] and [A1]. The larger-cutoff average is a mixture of the first average and the average over the added interval; the latter is at least the first whenever it has positive mass, and if it has zero mass the two averages agree.

By [F3], $P(A_k)\le c_k=kq\mathrm{/}m$ for every $k$. Substituting into step 1.1 yields $\mathbb{E}[I_i\mathrm{/}(R\vee 1)]\le (q\mathrm{/}m){\sum }_{k=1}^m{a}_k$.

Let $k_0$ be the first index with $P(A_{k_0})>0$, if one exists; then $a_j=0$ for $j<k_0$, and $C_{i,k_0}\subseteq D_{i,k_0}$ gives $a_{k_0}\le P(P_{-i}\in D_{i,k_0}\mid A_{k_0})\le 1$ by step 2.1. For $k>k_0$, suppose ${\sum }_{j<k}{a}_j\le P(P_{-i}\in D_{i,k-1}\mid A_{k-1})$. Since $P(A_{k-1})>0$ and $A_{k-1}\subseteq A_k$, step 2.1 applied to the nondecreasing $g_{k-1}$ makes this at most $P(P_{-i}\in D_{i,k-1}\mid A_k)$. Also $P(A_k)>0$, so [F4] gives $a_k=P(P_{-i}\in C_{i,k}\mid A_k)$. The disjoint union $D_{i,k}=D_{i,k-1}\dot{\cup }{C}_{i,k}$ from [F2] then yields ${\sum }_{j\le k}{a}_j\le P(P_{-i}\in D_{i,k}\mid A_k)$, completing the induction.

If $k_0$ exists, step 3.1 at $k=m$ and $D_{i,m}$ equal to the whole space give ${\sum }_{k=1}^m{a}_k\le 1$. If no such $k_0$ exists, all $P(A_k)=0$ and every $a_k=0$, so the same bound holds. Substituting into step 2.2 proves the contribution inequality.

Fix $\theta \in \mathrm{\Theta }$. Let $I_i$ be the indicator that the BH rule [F2] rejects hypothesis $i$. By [F1], $V_{\theta }={\sum }_{i\in I_0(\theta )}{I}_i$ and ${\mathrm{FDR}}_{\theta }=E_{\theta }[V_{\theta }\mathrm{/}(R\vee 1)]$. The ratio is bounded by one, so it is integrable by [F6]; finite linearity [F5] gives ${\mathrm{FDR}}_{\theta }={\sum }_{i\in I_0(\theta )}{E}_{\theta }[I_i\mathrm{/}(R\vee 1)]$.

If $I_0(\theta )=\mathrm{\varnothing }$, step 1.1 is an empty sum and FDR is zero. Otherwise, for each $i\in I_0(\theta )$, [A1] gives validity and [A2] gives PRDS on a set containing $i$ under the AC-based conditional-law interface [A3], so [F4] bounds its contribution by $q\mathrm{/}m$. Summing step 1.1 gives ${\mathrm{FDR}}_{\theta }\le \mathrm{\mid }{I}_0(\theta )\mathrm{\mid }q\mathrm{/}m=(m_0(\theta )\mathrm{/}m)q\le q$. Only the designated true-null coordinates are needed; independence is not assumed.