Proof of Good-Bad Splitting of the Integrated Symmetrised Score Functional: Chebyshev Bound for the Under-Likelihood Set, Transfer of Mass Between Densities, and the Split Bound

Throughout, sums, differences, products and scalar multiples of measurable real-valued maps are measurable by claim 2 and claim 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, indicators of measurable sets are measurable by its claim 1, and integrals of finite sums of nonnegative measurable maps are the sums of the integrals, and the integral is monotone, by Linearity and Monotonicity of the Lebesgue Integral. We write $\mathsf{N}_w(a,b)=\sum_{q=1}^{n}w_q(a-b_q)$ as in Weighted Cauchy-Schwarz Inequality on a Measure Space and the Symmetrised Score Functional: Bounds and Averaging, and $\mathsf{U}_q=\{z:\mathsf{r}_qL_q(z)<1-\delta\}$, so that $\mathsf{U}=\bigcup_{q=1}^{n}\mathsf{U}_q$ and $\pi_q=\int_{\mathsf{U}_q}\ell\,d\nu$.

Step 1 (claim 1). Measurability. The map $t\mapsto1/t$ is sequentially continuous on $(0,\infty)$ (by Arithmetic of Limits of Real Sequences), and $\ell$ takes values in $(0,\infty)$, so $1/\ell$ is measurable by Sequentially Continuous Functions of Measurable Euclidean Maps are Measurable (with $E=(0,\infty)$), and $L_q=\ell_q\cdot(1/\ell)$ is measurable as a product. Hence each $\mathsf{U}_q=\{\mathsf{r}_qL_q-(1-\delta)<0\}$ is measurable and so is the finite union $\mathsf{U}$. Integrals. $\ell L_q=\ell_q$ pointwise, so $\int\ell L_q\,d\nu=1$. Pointwise, $\ell(L_q-1)^{2}+2\ell L_q=\ell L_q^{2}+\ell$ (all terms nonnegative), so by additivity $V_q+2=\int\ell L_q^{2}\,d\nu+1$ in $[0,\infty]$, i.e. $\int\ell L_q^{2}\,d\nu=1+V_q$. Thus $V_q<\infty$ if and only if $\int\ell L_q^{2}\,d\nu<\infty$, and then $V_q=\int\ell L_q^{2}\,d\nu-1$.

Step 2 (claim 2). On $\mathsf{U}_q$ one has $1-\mathsf{r}_qL_q>\delta>0$, hence $(1-\mathsf{r}_qL_q)^{2}>\delta^{2}$, and therefore $\mathbf{1}_{\mathsf{U}_q}\le\delta^{-2}(1-\mathsf{r}_qL_q)^{2}$ pointwise on $\mathsf{Z}$ (the right side being nonnegative off $\mathsf{U}_q$). Multiplying by $\ell$ and integrating, $\pi_q\le\delta^{-2}\int\ell(1-\mathsf{r}_qL_q)^{2}\,d\nu$. If $\mathsf{r}_q=0$, the integral is $\int\ell\,d\nu=1=(1-\mathsf{r}_q)^{2}+\mathsf{r}_q^{2}V_q$ (with $0\cdot\infty=0$), and the claim follows. If $\mathsf{r}_q>0$ and $V_q=\infty$, the right side of the claim is $\infty$ and there is nothing to prove. If $\mathsf{r}_q>0$ and $V_q<\infty$, then $\int\ell L_q^{2}\,d\nu<\infty$ by Step 1; the pointwise identity $\ell(1-\mathsf{r}_qL_q)^{2}+2\mathsf{r}_q\ell L_q=\ell+\mathsf{r}_q^{2}\ell L_q^{2}$ between nonnegative measurable maps, integrated by additivity, shows that both sides have the finite total $1+\mathsf{r}_q^{2}(1+V_q)$, and hence $\int\ell(1-\mathsf{r}_qL_q)^{2}\,d\nu=1+\mathsf{r}_q^{2}(1+V_q)-2\mathsf{r}_q=(1-\mathsf{r}_q)^{2}+\mathsf{r}_q^{2}V_q$.

Step 3 (claim 3). Fix $A\in\mathcal{Z}$ and $q$, and put $I_A=\int_A\ell\,d\nu\in[0,1]$.

Case $V_q<\infty$. Apply claim 1 of Weighted Cauchy-Schwarz Inequality on a Measure Space and the Symmetrised Score Functional: Bounds and Averaging on $(\mathsf{Z},\mathcal{Z},\nu)$ with $\beta=\ell\mathbf{1}_A$ (measurable, $\int\beta\,d\nu=I_A<\infty$) and $\alpha=\ell(L_q-1)\mathbf{1}_A$ (measurable, vanishing where $\beta=0$, i.e. off $A$, since $\ell>0$). The associated map $q_\alpha$ (the map called $q$ in that claim, renamed here since $q$ is an index), equal to $\alpha^{2}/\beta$ on $\{\beta>0\}=A$ and to $0$ off $A$, equals $\ell(L_q-1)^{2}\mathbf{1}_A\le\ell(L_q-1)^{2}$, so $\int q_\alpha\,d\nu\le V_q<\infty$; the claim yields that $\alpha$ is integrable and $(\int\alpha\,d\nu)^{2}\le I_AV_q$, hence $\int\alpha\,d\nu\le(I_AV_q)^{1/2}=V_q^{1/2}I_A^{1/2}$ (a real number $s$ with $s^{2}\le c$ satisfies $s\le c^{1/2}$, since $s>c^{1/2}\ge0$ would give $s^{2}>c$; and $(st)^{1/2}=s^{1/2}t^{1/2}$ for $s,t\ge0$ by uniqueness of the nonnegative square root, Existence and Uniqueness of the Nonnegative Square Root). Now $\ell_q\mathbf{1}_A=\ell\mathbf{1}_A+\alpha$ pointwise, where $\ell_q\mathbf{1}_A$ and $\ell\mathbf{1}_A$ are nonnegative with integrals at most $1$, hence integrable; by the linearity of the integral for integrable maps (claim 2 of Linearity and Monotonicity of the Lebesgue Integral), $\int_A\ell_q\,d\nu=I_A+\int\alpha\,d\nu\le I_A+V_q^{1/2}I_A^{1/2}$.

Case $V_q=\infty$. If $I_A>0$ the right side is $\infty$ and there is nothing to prove. If $I_A=0$, then for every natural number $k$, $\nu(A\cap\{\ell>1/k\})=\int\mathbf{1}_{A\cap\{\ell>1/k\}}\,d\nu\le k\int_{A\cap\{\ell>1/k\}}\ell\,d\nu\le kI_A=0$ by the integral of an indicator (Simple Function and Its Integral) and monotonicity (as $\mathbf{1}_{A\cap\{\ell>1/k\}}\le k\ell\mathbf{1}_{A\cap\{\ell>1/k\}}$); since $\ell>0$, $A$ is the countable union of these sets, so $\nu(A)=0$ by countable subadditivity (claim 4 of Basic Properties of a Measure). Then $\int\min(\ell_q,k)\mathbf{1}_A\,d\nu\le k\nu(A)=0$ for every $k$ (the map $\min(\ell_q,k)$ being measurable by claim 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions), and $\min(\ell_q,k)\mathbf{1}_A$ increases pointwise to $\ell_q\mathbf{1}_A$, so $\int_A\ell_q\,d\nu=0$ by Monotone Convergence Theorem, which is the claimed inequality with right side $0+\infty\cdot0=0$.

Step 4 (claim 4). Measurability. The maps $\ell$ and $\mathsf{r}_q\ell_q$ are measurable, $[0,\infty)$-valued, with finite integrals $1$ and $\mathsf{r}_q$; claim 3 of Weighted Cauchy-Schwarz Inequality on a Measure Space and the Symmetrised Score Functional: Bounds and Averaging (its measurability assertion) shows that $z\mapsto\Phi_w(\ell(z),\mathsf{r}_1\ell_1(z),\dots,\mathsf{r}_n\ell_n(z))$ is measurable. The map $\ell(\sum_qw_q(1-\mathsf{r}_qL_q))^{2}$ is measurable by the arithmetic rules.

Pointwise bound off $\mathsf{U}$. Let $z\notin\mathsf{U}$. Then $\mathsf{r}_q\ell_q(z)\ge(1-\delta)\ell(z)$ for every $q$ (as $\ell(z)>0$ and $\mathsf{r}_qL_q(z)\ge1-\delta$), and $\ell(z)>0$, so claim 2(d) of Weighted Cauchy-Schwarz Inequality on a Measure Space and the Symmetrised Score Functional: Bounds and Averaging with $a=\ell(z)$ and $b_q=\mathsf{r}_q\ell_q(z)$ gives $\Phi_w(\ell(z),(\mathsf{r}_q\ell_q(z))_q)\le(1+\delta)\mathsf{N}_w(a,b)^{2}/\ell(z)$, where $\mathsf{N}_w(a,b)=\sum_qw_q(\ell(z)-\mathsf{r}_q\ell_q(z))=\ell(z)\sum_qw_q(1-\mathsf{r}_qL_q(z))$; hence $\Phi_w(\dots)\le(1+\delta)\,\ell(z)\bigl(\sum_qw_q(1-\mathsf{r}_qL_q(z))\bigr)^{2}$.

Pointwise bound on $\mathsf{U}$. For every $z$, claim 2(b) of Weighted Cauchy-Schwarz Inequality on a Measure Space and the Symmetrised Score Functional: Bounds and Averaging gives $\Phi_w(\ell(z),(\mathsf{r}_q\ell_q(z))_q)\le2n\lVert w\rVert_1^{2}\bigl(\ell(z)+\sum_q\mathsf{r}_q\ell_q(z)\bigr)$.

Integration. Writing $\Phi_w(\dots)=\mathbf{1}_{\mathsf{Z}\setminus\mathsf{U}}\Phi_w(\dots)+\mathbf{1}_{\mathsf{U}}\Phi_w(\dots)$ and using the two pointwise bounds, additivity and monotonicity, $\int_{\mathsf{Z}}\Phi_w(\ell,\mathsf{r}_1\ell_1,\dots,\mathsf{r}_n\ell_n)\,d\nu\le(1+\delta)\int_{\mathsf{Z}\setminus\mathsf{U}}\ell\Bigl(\sum_qw_q(1-\mathsf{r}_qL_q)\Bigr)^{2}d\nu+2n\lVert w\rVert_1^{2}\Bigl[\int_{\mathsf{U}}\ell\,d\nu+\sum_q\mathsf{r}_q\int_{\mathsf{U}}\ell_q\,d\nu\Bigr].$ The first integral is at most the same integral over $\mathsf{Z}$ (monotonicity). For the bracket: $\mathbf{1}_{\mathsf{U}}\le\sum_q\mathbf{1}_{\mathsf{U}_q}$ pointwise, so $\int_{\mathsf{U}}\ell\,d\nu\le\sum_q\pi_q=\pi$; and by claim 3 with $A=\mathsf{U}$, $\int_{\mathsf{U}}\ell_q\,d\nu\le\int_{\mathsf{U}}\ell\,d\nu+V_q^{1/2}(\int_{\mathsf{U}}\ell\,d\nu)^{1/2}\le\pi+V_q^{1/2}\pi^{1/2}$, the square root being nondecreasing on $[0,\infty)$ (Existence and Uniqueness of the Nonnegative Square Root: if $0\le s\le t$ then $s^{1/2}\le t^{1/2}$, since otherwise squaring would give $s>t$). Substituting gives the split bound, together with the two refinements stated after it. $\blacksquare$