Loss, Risk and Decision Theory

If $\int fd\mu =+\mathrm{\infty }$, the conclusion holds because $f(b)\in [0,+\mathrm{\infty }]$. Otherwise let $A:=\{a\in C:f(a)=+\mathrm{\infty }\}$. For every integer $n\ge 1$, the simple function $n{\mathbf{1}}_A$ is bounded above by $f$, so [F7] gives $n\mu (A)=\int n{\mathbf{1}}_{A}d\mu \le \int fd\mu <\mathrm{\infty }$. If $\mu (A)>0$, these lower bounds are unbounded as $n\to \mathrm{\infty }$ (and are already infinite if $\mu (A)=+\mathrm{\infty }$), a contradiction. Thus $E:=C\setminus A=\{a:f(a)<\mathrm{\infty }\}$ has full $\mu$-measure. Let $S$ be the support of $\mu$, the set of points every relative open neighbourhood of which has positive measure. Since $C$ has a countable base inherited from rational balls in ${\mathbb{R}}^d$, the complement of $S$ is a countable union of open null basic sets, so $\mu (S)=1$. Every neighbourhood of a point of $S$ meets $E$, hence $E\cap S$ is dense in $S$.

Let $C_0$ be the closure of the set of finite convex combinations of points of $S$. Then $C_0\subseteq C$. If $b\mathrm{\notin }{C}_0$, [F1] gives a linear functional $u$ and a real $\alpha$ with $⟨u,y⟩\le \alpha <⟨u,b⟩$ for all $y\in C_0$. The function $\psi (a)=\alpha -⟨u,a⟩$ is integrable by the first-moment assumption and is nonnegative on the full-measure support $S$. Hence ${\psi }^{-}=0$ almost everywhere, so [F5] gives $\int {\psi }^{-}d\mu =0$ and $\int \psi d\mu =\int {\psi }^{+}d\mu \ge 0$. By [F6] and the coordinate definition of $b$, this says $\alpha -⟨u,b⟩\ge 0$, contradicting the strict separation. Thus $b\in C_0$.

If $b$ were on the relative boundary of $C_0$, [F2], applied in its affine hull, would give a nonzero supporting normal $v$ there with $⟨v,y-b⟩\le 0$ for all $y\in C_0$. Set $u(a)=⟨v,b-a⟩$. The first-moment assumption makes $u$ integrable, and [F6] gives $\int ud\mu =0$. Since $u\ge 0$ on the full-measure support $S$, its negative part $u^{-}$ vanishes almost everywhere; [F5] gives $\int u^{-}d\mu =0$. The signed-integral definition in [F6] then gives $\int u^{+}d\mu =0$, so [F5] implies $u^{+}=0$ almost everywhere and therefore $u=0$ almost everywhere. By continuity of $u$ and the definition of support, every point of $S$ must lie in the hyperplane $⟨v,y-b⟩=0$: otherwise a neighborhood of a support point outside it would have positive measure on which $u\mathrm{\ne }0$. Its closed convex hull $C_0$ would lie in that hyperplane, contradicting that $v$ is nonzero in the direction space of its affine hull. Hence $b$ is in the relative interior of $C_0$.

Put $C_1=\mathrm{conv}(E\cap S)$, where $\mathrm{conv}$ denotes finite convex combinations. The extended-functional convexity inequality makes $f$ finite on $C_1$; on these finite values it is the real-valued Euclidean convexity of Convex and strictly convex functions on Euclidean convex sets. Density from step 1.1 gives $\overline{C_1}=C_0$. By [F3], $b$ is in the relative interior of $C_1$. If this affine hull has dimension zero, then $\mu$ is concentrated at $b$ and the claim is equality. Otherwise identify the affine hull isometrically with ${\mathbb{R}}^k$; [F4] applied to $f{\mathrm{\mid }}_{C_1}$ gives a subgradient $v$ at $b$, so $f(y)\ge f(b)+⟨v,y-b⟩$ for every $y\in C_1$.

Let $\mathrm{\ell }(a)=f(b)+⟨v,a-b⟩$ and $w=(f-\mathrm{\ell }){\mathbf{1}}_E$, defining $w=0$ on $E^c$. The function $w$ is integrable because $f{\mathbf{1}}_E$ has finite nonnegative integral and $\mathrm{\ell }$ is integrable by the first-moment assumption. The supporting inequality gives $w\ge 0$ on $E\cap S$, a full-measure set; thus $w^{-}=0$ almost everywhere. By [F5], $\int w^{-}d\mu =0$, so [F6] gives $\int wd\mu =\int w^{+}d\mu \ge 0$. Linearity [F6] now yields $\int f{\mathbf{1}}_{E}d\mu \ge \int \mathrm{\ell }{\mathbf{1}}_{E}d\mu$. Also $f=f{\mathbf{1}}_E+f{\mathbf{1}}_{E^c}$, with $f{\mathbf{1}}_{E^c}=0$ almost everywhere; [F5] and additivity [F8] give $\int fd\mu =\int f{\mathbf{1}}_{E}d\mu$. The integrable function $\mathrm{\ell }{\mathbf{1}}_{E^c}$ is likewise zero almost everywhere, so [F5] applied to its positive and negative parts and linearity [F6] give $\int \mathrm{\ell }d\mu =\int \mathrm{\ell }{\mathbf{1}}_{E}d\mu$. Finally [F6] and the barycenter definition yield $\int \mathrm{\ell }d\mu =f(b)+⟨v,b-b⟩=f(b)$. This proves the claim in the finite-integral case; the infinite-integral case was handled in step 1.1.

For each $C\in \mathcal{T}\otimes \mathcal{A}$, [F2] makes $C_{\theta }$ measurable and [F3], applied to $1_C$, makes $\theta \mapsto K(\theta ,C_{\theta })$ measurable. Thus $J(C)$ is defined. The empty set has $J$-mass zero, while $(\mathrm{\Theta }\times \mathcal{X}{)}_{\theta }=\mathcal{X}$ and [F1], [F5] give $J(\mathrm{\Theta }\times \mathcal{X})=1$.

If $(C_n{)}_{n\ge 0}$ are pairwise disjoint product-measurable sets, their sections are pairwise disjoint. Countable additivity of each $K(\theta ,\cdot )$ gives $K(\theta ,({\bigcup }_n{C}_n{)}_{\theta })={\lim }_N{\sum }_{n=0}^{N}K(\theta ,(C_n{)}_{\theta })$. The partial sums increase; [F4] passes this limit through the $\pi$-integral and yields $J({\bigcup }_n{C}_n)={\sum }_{n}J(C_n)$. Therefore $J$ is a probability measure.

For a rectangle $A\times B$, its section is $B$ on $A$ and empty off $A$, so the definition gives $J(A\times B)={\int }_{A}K(\theta ,B)\pi (d\theta )$. For a nonnegative simple function $h={\sum }_{i=1}^m{c}_i{1}_{C_i}$ with disjoint measurable sets and $c_i\ge 0$, the set formula for $J$ and [F6] give the displayed iterated-integral identity by finite addition and homogeneity.

For general nonnegative product-measurable $h$, take the increasing simple approximants of [F4]. Apply monotone convergence first under $J$, then sectionwise under each $K(\theta ,\cdot )$, and finally under $\pi$. The identity for the simple functions from step 3.1 passes to the limit, including when either side is $+\mathrm{\infty }$.

Let $m(x)={\int }_{\mathsf{D}}\mathrm{\parallel }a{\mathrm{\parallel }}_2\delta (x,da)$. By [F2], $m$ is measurable. For each coordinate $j$, $\mathrm{\mid }{a}_j\mathrm{\mid }\le \mathrm{\parallel }a{\mathrm{\parallel }}_2$, so the signed integral $b_j(x)=\int a_j\delta (x,da)$ is measurable on $E_0=\{x:m(x)<\mathrm{\infty }\}$ by [F2]; hence $b=(b_1,\dots ,b_d)$ is measurable on $E_0$.

The set $E=\{x\in E_0:b(x)\in \mathsf{D}\}$ is measurable because $\mathsf{D}$ is closed and therefore Borel. Fix one $a_0\in \mathsf{D}$ and define $d(x)=b(x)$ on $E$ and $d(x)=a_0$ off $E$. This is a measurable map into $\mathsf{D}$; the hypotheses give $P_{\theta }(E)=1$ for every $\theta$.

For each $\theta$ and every $x\in E$, the section $a\mapsto L(\theta ,a)$ is nonnegative Borel convex, $\delta (x,\cdot )$ is a probability measure with finite first moment, and its barycenter $b(x)$ belongs to $\mathsf{D}$. Applying [F3] gives $L(\theta ,d(x))=L(\theta ,b(x))\le {\int }_{\mathsf{D}}L(\theta ,a)\delta (x,da)$.

Put $g_{\theta }(x)=L(\theta ,d(x))$ and $h_{\theta }(x)={\int }_{\mathsf{D}}L(\theta ,a)\delta (x,da)$. Both are nonnegative measurable functions by [F1], [F2], and measurability of $d$. Step 2.1 gives $g_{\theta }{1}_E\le h_{\theta }{1}_E$, so monotonicity [F5] gives $\int g_{\theta }{1}_{E}d{P}_{\theta }\le \int h_{\theta }{1}_{E}d{P}_{\theta }$. Since $P_{\theta }(E^c)=0$, [F4] gives zero integrals on $E^c$; additivity [F5] then yields $\int g_{\theta }d{P}_{\theta }\le \int h_{\theta }d{P}_{\theta }$. This is $R(\theta ,d)\le R(\theta ,\delta )$, and it holds for every $\theta$.

Fix a randomized rule $\delta$ and put $h_{\delta }(\theta ,x)={\int }_{\mathsf{D}}L(\theta ,a)\delta (x,da)$. The lifted kernel $((\theta ,x),B)\mapsto \delta (x,B)$ is a probability kernel because the coordinate projection $(\theta ,x)\mapsto x$ is measurable by [F12]. The lifted integrand $(\theta ,x,a)\mapsto L(\theta ,a)$ is product-measurable because the pairing map $((\theta ,x),a)\mapsto (\theta ,a)$ is measurable by [F12]. By [F5], $h_{\delta }$ is product-measurable. The risk definition [F2], prior-risk definition [F11], and joint-law identity [F3] therefore give $r_{\pi }(\delta )=\int h_{\delta }dJ$.

Apply the posterior disintegration identity [F4] to $h_{\delta }\ge 0$. If $M$ is the second marginal of $J$, then $r_{\pi }(\delta )={\int }_{\mathcal{X}}({\int }_{\mathrm{\Theta }}{h}_{\delta }(\theta ,x){\mathrm{\Pi }}_x(d\theta ))M(dx)$. Every integral is an extended nonnegative integral, so no integrability beyond the stated measurability is needed.

Lift $\mathrm{\Pi }$ to the kernel $((x,a),B)\mapsto {\mathrm{\Pi }}_x(B)$ from $\mathcal{X}\times \mathsf{D}$ to $\mathrm{\Theta }$. The pairing map $((x,a),\theta )\mapsto (\theta ,a)$ is measurable by [F12], so [F5] makes $q(x,a)$ product-measurable. For each fixed $x$, [F6] applied to the product probability ${\mathrm{\Pi }}_x\times \delta (x,\cdot )$ and $L(\theta ,a)$ gives ${\int }_{\mathrm{\Theta }}{h}_{\delta }(\theta ,x){\mathrm{\Pi }}_x(d\theta )={\int }_{\mathsf{D}}q(x,a)\delta (x,da)$. The right side is measurable in $x$ by [F5]. Thus step 2.1 yields $r_{\pi }(\delta )={\int }_{\mathcal{X}}{\int }_{\mathsf{D}}q(x,a)\delta (x,da)M(dx)$.

The point-mass kernel ${\delta }^{\ast }(x,B)=1_B(d^{\ast }(x))$ is measurable because $d^{\ast }$ is measurable by [F10]; applying step 3.1 gives $r_{\pi }(d^{\ast })=\int gdM$ for $g(x)=q(x,d^{\ast }(x))$. The pairing map $x\mapsto (x,d^{\ast }(x))$ is measurable by [F10, F12], so $g$ is measurable. Put $h(x)=\int q(x,a)\delta (x,da)$, which is measurable by [F5]. On the measurable full-measure set $E$ from the statement, $g{1}_E\le h{1}_E$, so [F8] gives $\int g{1}_{E}dM\le \int h{1}_{E}dM$. The complementary integrals vanish by [F7]; additivity [F9] gives $\int gdM\le \int hdM=r_{\pi }(\delta )$ by step 3.1. The rule $\delta$ was arbitrary.