Integral Jensen inequality for finite-dimensional convex losses

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.