Neyman–Pearson lemma with singular alternative mass

First suppose $\alpha >0$, so the cutoff $c$ in [F5] is finite. Put $D={\phi }_{\alpha }^{\ast }-\phi$. Then $D$ is measurable and $\mathrm{\mid }D\mathrm{\mid }\le 1$. By [F2] and [F6], $\int p_{i}d\mu =P_i(\mathcal{X})=1$, so $\mathrm{\mid }D{p}_i\mathrm{\mid }\le p_i$ makes $D{p}_i$ integrable for $i=0,1$. Since $c$ is finite, $D(p_1-c{p}_0)$ is also integrable.

On $\{p_0=0\}$, if $p_1>0$ then [F5] gives ${\phi }_{\alpha }^{\ast }=1$, so $D(p_1-c{p}_0)\ge 0$; if $p_1=0$, the product is zero. On $\{p_0>0\}$, write $R=p_1\mathrm{/}{p}_0$. If $R>c$, then ${\phi }_{\alpha }^{\ast }=1$ and both $D$ and $p_1-c{p}_0$ are nonnegative. If $R<c$, then ${\phi }_{\alpha }^{\ast }=0$ and both are nonpositive. If $R=c$, the second factor is zero. Thus $D(p_1-c{p}_0)\ge 0$ everywhere.

The signed density identity [F15] and linearity [F8] give $\int D(p_1-c{p}_0)d\mu =(E_{P_1}{\phi }_{\alpha }^{\ast }-E_{P_1}\phi )-c(E_{P_0}{\phi }_{\alpha }^{\ast }-E_{P_0}\phi )$.

By monotonicity of the nonnegative integral, the integral in step 2.2 is nonnegative. Hence $E_{P_1}{\phi }_{\alpha }^{\ast }-E_{P_1}\phi \ge c(E_{P_0}{\phi }_{\alpha }^{\ast }-E_{P_0}\phi )$. The candidate has null rejection probability $\alpha$ by [F5], while the competitor has null rejection probability at most $\alpha$ by [F4]. Since $c\ge 0$, the right-hand side is nonnegative, proving the desired power inequality for $\alpha >0$.

Now suppose $\alpha =0$. Then $E_{P_0}\phi =0$, so [F6] gives $\int \phi p_{0}d\mu =0$. For positive integers $n,m$, set $B_{n,m}=\{\phi \ge 1\mathrm{/}n\}\cap \{p_0\ge 1\mathrm{/}m\}$. On this set, $\phi p_0\ge 1\mathrm{/}(nm)$; [F7] and [F11] therefore give $0\ge {\int }_{B_{n,m}}\phi p_{0}d\mu \ge \mu (B_{n,m})\mathrm{/}(nm)$. Thus $\mu (B_{n,m})=0$, and absolute continuity [F1] gives $P_1(B_{n,m})=0$. For each positive-integer pair $(n,m)$ put $C_{n,m}=\{\phi \ge 1\mathrm{/}n\}\cap \{p_0\ge 1\mathrm{/}m\}$, enumerate these pairs by increasing $n+m$ and then by increasing $n$, and write the resulting sequence of sets as $(C_k{)}_{k\ge 0}$. If $\phi (x)>0$ and $p_0(x)>0$, [F16] supplies positive integers $n,m$ with $1\mathrm{/}n<\phi (x)$ and $1\mathrm{/}m<p_0(x)$, so $x\in {\bigcup }_k{C}_k$. Conversely every $C_k$ is contained in $\{\phi >0,p_0>0\}$; hence this union is exactly that set. Put $D_0=C_0$ and $D_k=C_k\setminus {\bigcup }_{j<k}{C}_j$ for $k\ge 1$. The $D_k$ are measurable by [F13], pairwise disjoint, and each $D_k\subseteq C_k$, so the preceding null-set calculation and [F12] give $P_1(D_k)=0$. Their union equals ${\bigcup }_k{C}_k$: if $x$ lies in that union, the nonempty set of indices $k$ with $x\in C_k$ has a least member by [F14], which puts $x$ in the corresponding $D_k$. Countable additivity [F9] makes the union $P_1$-null. On its complement, $\phi =0$ wherever $p_0>0$; where $p_0=0<p_1$, [F5] gives ${\phi }_0^{\ast }=1$, and the remaining set $\{p_0=p_1=0\}$ has $P_1$ measure zero because $p_1=0$ there and [F2], [F6] give $P_1(A)={\int }_A{p}_{1}d\mu$. Therefore $\phi \le {\phi }_0^{\ast }$ $P_1$-almost everywhere, and monotonicity of expectation proves the power inequality at $\alpha =0$.