Fixed-bin calibration limits underestimate for jointly convex discrepancies; refining bins can be consistent

Fix a partition $\mathcal{P}$. For every bin $B$ and each coordinate, the variables ${\mathbf{1}}_{\{G_i\in B\}}$, ${\mathbf{1}}_{\{G_i\in B\}}(V_i{)}_j$, and ${\mathbf{1}}_{\{G_i\in B\}}(G_i{)}_j$ are bounded iid real variables: the same measurable coordinate map is applied to each iid pair, so the outputs have the same law by taking preimages, and their independence follows from [F3]. Their sample averages therefore converge almost surely to their expectations. On the event $P(G\in B)>0$, division by $N_B\mathrm{/}n\to P(G\in B)$ gives convergence of the bin means to $\mathbb{E}[V\mid G\in B]$ and $\mathbb{E}[G\mid G\in B]$. On a zero-probability bin every sample has count zero almost surely. Since the partition and label set are finite, these convergences hold together almost surely; continuity of $d$ then gives ${\widehat{E}}_{n,\mathcal{P}}\to E_{\mathcal{P}}$. By [F1], $\mathbb{E}[V\mid G\in B]=\mathbb{E}[r(G)\mid G\in B]$ coordinatewise.

Let ${\delta }_m$ be the maximum Euclidean diameter of a cell of ${\mathcal{P}}_m$. For any fixed $v\in {\mathrm{\Delta }}_K$, the function $\psi (u,w)=\mathrm{\parallel }u-w{\mathrm{\parallel }}_2$ is continuous and jointly convex: the metric triangle inequality gives continuity in $(u,w)$, and $\mathrm{\parallel }\lambda (u-w)+(1-\lambda )(u^{\mathrm{\prime }}-w^{\mathrm{\prime }}){\mathrm{\parallel }}_2\le \lambda \mathrm{\parallel }u-w{\mathrm{\parallel }}_2+(1-\lambda )\mathrm{\parallel }{u}^{\mathrm{\prime }}-w^{\mathrm{\prime }}{\mathrm{\parallel }}_2$ for $0\le \lambda \le 1$ by the triangle inequality and homogeneity in [F5]. For a positive-probability bin $B$, define $P_B(A)=P(A\cap \{G\in B\})\mathrm{/}P(G\in B)$. If $G(\omega )\in B$, apply [F2] under $P_B$ to the simplex pair $(G,G(\omega ))$. It gives $\mathrm{\parallel }\mathbb{E}[G\mid G\in B]-G(\omega ){\mathrm{\parallel }}_2\le \mathbb{E}[\mathrm{\parallel }G-G(\omega ){\mathrm{\parallel }}_2\mid G\in B]\le {\delta }_m$. Apply the same argument to $(r(G),r(G(\omega )))$ to obtain $\mathrm{\parallel }\mathbb{E}[r(G)\mid G\in B]-r(G(\omega )){\mathrm{\parallel }}_2\le L{\delta }_m$. Thus the Euclidean distance between the ordered pairs is at most $\sqrt{1+L^2}{\delta }_m\le (1+L){\delta }_m$. By [F4], $d$ is uniformly continuous on the compact product simplex, so the absolute difference between the bin value and $d(r(G(\omega )),G(\omega ))$ is bounded by a modulus tending to zero with ${\delta }_m$, uniformly over all bins and $\omega$. Integrating over bins yields $E_{{\mathcal{P}}_m}\to \mathbb{E}[d(r(G),G)]$.

For the underestimation conclusion assume additionally that $d$ is jointly convex. For each positive-probability bin define the conditional probability measure $P_B(A)=P(A\cap \{G\in B\})\mathrm{/}P(G\in B)$. Under $P_B$, the pair $(r(G),G)$ takes values in ${\mathrm{\Delta }}_K\times {\mathrm{\Delta }}_K$. Apply [F2] to $\psi =d$ and multiply by $P(G\in B)$ to obtain $P(G\in B)d\text{\hspace{0.17em}⁣}(\mathbb{E}[r(G)\mid G\in B],\mathbb{E}[G\mid G\in B])\le \mathbb{E}[{\mathbf{1}}_{\{G\in B\}}d(r(G),G)]\mathrm{.}$ Summing over the partition proves $E_{\mathcal{P}}\le \mathbb{E}[d(r(G),G)]$.

For each fixed $m$, step 1.1 gives the almost-sure limit in $n$. There are countably many $m$, so the intersection of these probability-one events still has probability one; step 1.2 then gives the iterated limit as stated. This order takes the sample-size limit before the partition refinement and makes no claim about a jointly chosen diagonal sequence.