Group-sequential boundaries exist uniquely and spend alpha exactly

For $1\le i\le j\le K$, define a lower-triangular matrix $L$ by $L_{j1}=\sqrt{t_1\mathrm{/}{t}_j}$ and, for $2\le i\le j$, $L_{ji}=\sqrt{(t_i-t_{i-1})\mathrm{/}{t}_j}$, with $L_{ji}=0$ for $i>j$. For $j\le k$, the $(j,k)$ entry of $L{L}^T$ is $\frac{t_1+{\sum }_{i=2}^j(t_i-t_{i-1})}{\sqrt{t_j{t}_k}}=\sqrt{t_j\mathrm{/}{t}_k}={\mathrm{\Sigma }}_{jk}\mathrm{.}$ Thus $\mathrm{\Sigma }=L{L}^T$. Its diagonal entries in $L$ are $1$ and $\sqrt{(t_i-t_{i-1})\mathrm{/}{t}_i}>0$ for $i\ge 2$, so $L$ is invertible and $v^T\mathrm{\Sigma }v=\mathrm{\parallel }{L}^{T}v{\mathrm{\parallel }}^2>0$ for every nonzero $v$.

For every $k$, the $k$th coordinate has law $N(0,1)$ by the projection clause in [F2]. It has no atoms, and for $x<y$, $0<P_0(x<Z_k<y)={\int }_x^y\varphi (u)du\le \frac{y-x}{\sqrt{2\pi }}\mathrm{.}$ Positivity follows from [F5]; the upper bound follows from $0<e^{-u^2\mathrm{/}2}\le 1$. The same facts bound a two-sided strip by $P_0(c<\mathrm{\mid }{Z}_k\mathrm{\mid }<d)\le 2(d-c)\mathrm{/}\sqrt{2\pi }$ when $0\le c<d$.

By [F2] and [F3], the canonical law exists and is represented by $Z={\mathrm{\Sigma }}^{1\mathrm{/}2}Y$ for independent standard-normal coordinates $Y$. Since $\mathrm{\Sigma }$ is positive definite by step 1.1, ${\mathrm{\Sigma }}^{1\mathrm{/}2}$ is invertible: if ${\mathrm{\Sigma }}^{1\mathrm{/}2}v=0$, then $v^T\mathrm{\Sigma }v=0$, forcing $v=0$. The inverse image under this linear homeomorphism of any nonempty open subset of ${\mathbb{R}}^K$ is nonempty and open, hence contains a product of nonempty open intervals. Each interval has positive standard-normal probability by [F4]–[F5], and independence makes the product probability positive. Therefore every nonempty open subset of ${\mathbb{R}}^K$ has positive $P_0$-probability.

For the induction step, fix $k\in \{1,\dots ,K\}$. Assume that the boundaries through $k-1$ have been constructed with $P_0(R_{k-1})=A(t_{k-1})$; for $k=1$ this is the empty base case $R_0=\mathrm{\varnothing }$, $A(t_0)=A(0)=0$. Any finite two-sided boundary already constructed is positive, as established when its root was obtained. Let $S=R_{k-1}^c$. In the one-sided mode, $S$ is the intersection of the open conditions $Z_j<b_j$ for earlier finite boundaries; in the symmetric two-sided mode, it is the intersection of $\mathrm{\mid }{Z}_j\mathrm{\mid }<c_j$ for earlier finite boundaries. These conditions describe a nonempty open set: choose each constrained coordinate below its upper boundary, or inside its interval $(-c_j,c_j)$, and ignore infinite boundaries. Thus $P_0(S)=1-A(t_{k-1})\ge 1-\alpha >0$.

In the one-sided mode set $q(x)=A(t_{k-1})+P_0(S\cap \{Z_k\ge x\})$ for $x\in \mathbb{R}$. If $x<y$, then $q(x)-q(y)=P_0(S\cap \{x\le Z_k<y\})>0$. Although $S\subseteq {\mathbb{R}}^K$, it restricts only coordinates $j<k$, as described in step 3.1. Thus $O_{x,y}:=S\cap \{z\in {\mathbb{R}}^K:x<z_k<y\}$ is nonempty and open: choose each earlier coordinate below its finite one-sided boundary (or arbitrarily when its boundary is $+\mathrm{\infty }$), choose $z_k\in (x,y)$, and choose the remaining coordinates arbitrarily. By step 2.1, $P_0(O_{x,y})>0$; the only part of $S\cap \{x\le Z_k<y\}$ omitted from $O_{x,y}$ lies on $Z_k=x$, which has probability zero by step 1.2. Also $\mathrm{\mid }q(x)-q(y)\mathrm{\mid }\le (y-x)\mathrm{/}\sqrt{2\pi }$ by step 1.2, so $q$ is continuous. Moreover $q(x)\to 1$ as $x\to -\mathrm{\infty }$ and $q(x)\to A(t_{k-1})$ as $x\to +\mathrm{\infty }$, since the omitted lower and upper normal tails tend to zero. For the symmetric two-sided mode set $q(c)=A(t_{k-1})+P_0(S\cap \{\mathrm{\mid }{Z}_k\mathrm{\mid }\ge c\})$ for $c\ge 0$. For $0\le c<d$, $q(c)-q(d)=P_0(S\cap \{c\le \mathrm{\mid }{Z}_k\mathrm{\mid }<d\})>0$. Here $S$ again restricts only coordinates $j<k$, so $O_{c,d}:=S\cap \{z\in {\mathbb{R}}^K:c<\mathrm{\mid }{z}_k\mathrm{\mid }<d\}$ is nonempty and open: set every earlier coordinate to $0$ (which satisfies every finite earlier symmetric no-crossing condition because its boundary is positive), choose $z_k=(c+d)\mathrm{/}2$, and choose the remaining coordinates arbitrarily. Step 2.1 gives $P_0(O_{c,d})>0$. The bound in step 1.2 makes $q$ continuous. Here $q(0)=1$ and $q(c)\to A(t_{k-1})$ as $c\to +\mathrm{\infty }$. The tail limits in both modes follow from [F6] applied to increasing lower or central intervals of the standard-normal coordinate.

Since $A$ is nondecreasing, either $A(t_k)=A(t_{k-1})$ or $A(t_{k-1})<A(t_k)\le \alpha <1$. In the flat case, [F9] assigns $+\mathrm{\infty }$, whose crossing event is empty, so the previous cumulative probability is unchanged. In the increasing case, the target lies strictly between the endpoint limits of $q$. The limits in step 4.1 give finite endpoints on which the continuous function takes values on opposite sides of $A(t_k)$; [F7] gives a root. Strict decrease in step 4.1 makes that finite root unique. The induction therefore defines every boundary through the finite maximum $K$. In the symmetric two-sided mode $q(0)=1>A(t_k)$, so the root is strictly positive.

The induction step is complete: at an increasing look the selected root gives $P_0(R_k)=A(t_k)$ by [F9]; at a flat look the event union is unchanged and $A(t_k)=A(t_{k-1})$. Induction from $R_0=\mathrm{\varnothing }$ proves the equality for every $k$. Stopping at the first crossing makes $R_k$ exactly the event of rejection by look $k$. At $k=K$, $t_K=1$ and $A(1)=\alpha$, so the terminal simple-null rejection probability is $\alpha$. For any embedding with a nonempty alternative and singleton null $\{{\theta }_0\}$ whose law is $P_0$, the rejection indicator is a measurable test function because $R_K$ is a finite union of Borel threshold events; [F8] then makes its size $\alpha$. AC is used in the supplied Gaussian law and representation [F2] and [F4], not in the finite root selection: each root is unique and there are only finitely many looks.