SlowGuess/NeurIPS_2023_OCR huggingface dataset:
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In this section of the supplementary material, we provide detailed proofs for all results in the main text.
A.1 Proofs for Section 2
We start with Proposition 2.1, whose proof is based on [26, Lemma 1.2].
Proof. of Proposition 2.1 For  $M \in \mathbb{N}_{+}$  define the McShane extension  $F_{M} : \mathcal{P}(X) \times Z \to \mathbb{R}$  by
$$
F _ {M} (\mu , z) = \inf  _ {\vec {x} \in X ^ {M}} f _ {M} (\vec {x}, z) + L _ {f} d _ {\mathrm {K R}} (\hat {\mu} [ \vec {x} ], \mu).
$$
Observe that  $F_{M}$  is well-defined (i.e.,  $\mathbb{R}$ -valued) since  $f_{M}(\cdot, z)$  and  $L_{f}d_{\mathrm{KR}}(\hat{\mu}[\cdot], \mu)$  are bounded for every  $z \in Z$  (since  $f_{M}$  and  $d_{\mathrm{KR}}(\hat{\mu}[\cdot], \mu)$  are continuous and  $\mathcal{P}(X)$  is compact, hence bounded).
Step 1  $F_{M}$  extends  $f_{M}$ , i.e., for all  $M \in \mathbb{N}_{+}$ ,  $\vec{x} \in X^{M}$  and  $z \in Z$  we have  $F_{M}(\hat{\mu}[\vec{x}], z) = f_{M}(\vec{x}, z)$ . To show this, let  $\vec{x} \in X^{M}$  and  $z \in Z$  be arbitrary and observe that by definition
$$
F _ {M} (\hat {\mu} [ \vec {x} ], z) = \inf  _ {\vec {x} ^ {\prime} \in X ^ {M}} f _ {M} (\vec {x} ^ {\prime}, z) + L _ {f} d _ {\mathrm {K R}} (\hat {\mu} [ \vec {x} ^ {\prime}, \hat {\mu} [ \vec {x} ]) \leq f _ {M} (\vec {x}, z) + L _ {f} d _ {\mathrm {K R}} (\hat {\mu} [ \vec {x} ], \hat {\mu} [ \vec {x} ]) = f _ {M} (\vec {x}, z).
$$
If  $F_{M}(\hat{\mu}[\vec{x}], z) < f_{M}(\vec{x}, z)$ , then there exists some  $\vec{x}' \in X^{M}$  such that
$$
f _ {M} \left(\vec {x} ^ {\prime}, z\right) + L _ {f} d _ {\mathrm {K R}} \left(\hat {\mu} \left[ \vec {x} ^ {\prime} \right], \hat {\mu} [ \vec {x} ]\right) <   f _ {M} (\vec {x}, z),
$$
but this means that
$$
L _ {f} d _ {\mathrm {K R}} \left(\hat {\mu} [ \vec {x} ^ {\prime}, \hat {\mu} [ \vec {x} ]\right) <   f _ {M} (\vec {x}, z) - f _ {M} \left(\vec {x} ^ {\prime}, z\right) \leq \left| f _ {M} (\vec {x}, z) - f _ {M} \left(\vec {x} ^ {\prime}, z\right) \right|,
$$
contradicting the  $L_{f}$ -Lipschitz continuity of  $f_{M}$ .
Step 2 All  $F_{M}$  are  $L_{f}$ -continuous: Let  $M \in \mathbb{N}_{+}$ ,  $\mu_{i} \in \mathcal{P}(X)$  and  $z_{i} \in Z$ ,  $i = 1,2$ , be arbitrary. Since  $X^{M}$  is compact and  $f_{M}(\cdot ,z)$  and  $L_{f}d_{\mathrm{KR}}(\hat{\mu} [\cdot ],\mu_{i})$ ,  $i = 1,2$ , are continuous, the infimum in the definition of  $F_{M}$  is actually attained. Let  $\vec{x}_2 \in X^M$  such that  $F_{M}(\mu_{2},z_{2}) = f_{M}(\vec{x}_{2},z_{2}) + L_{f}d_{\mathrm{KR}}(\hat{\mu} [\vec{x}_{2}],\mu_{2})$ , then we have
$$
\begin{array}{l} F _ {M} \left(\mu_ {1}, z _ {1}\right) \leq f _ {M} \left(\vec {x} _ {2}, z _ {1}\right) + L _ {f} d _ {\mathrm {K R}} \left(\hat {\mu} [ \vec {x} _ {2} ], \mu_ {1}\right) \\ = f _ {M} \left(\vec {x} _ {2}, z _ {1}\right) + L _ {f} d _ {\mathrm {K R}} \left(\hat {\mu} \left[ \vec {x} _ {2} \right], \mu_ {2}\right) - L _ {f} d _ {\mathrm {K R}} \left(\hat {\mu} \left[ \vec {x} _ {2} \right], \mu_ {2}\right) + L _ {f} d _ {\mathrm {K R}} \left(\hat {\mu} \left[ \vec {x} _ {2} \right], \mu_ {1}\right) \\ \leq f _ {M} \left(\vec {x} _ {2}, z _ {2}\right) + L _ {f} d _ {\mathrm {K R}} \left(\hat {\mu} \left[ \vec {x} _ {2} \right], \mu_ {2}\right) + L _ {f} d _ {Z} \left(z _ {1}, z _ {2}\right) - L _ {f} d _ {\mathrm {K R}} \left(\hat {\mu} \left[ \vec {x} _ {2} \right], \mu_ {2}\right) \\ + L _ {f} d _ {\mathrm {K R}} \left(\hat {\mu} [ \vec {x} _ {2} ], \mu_ {1}\right) \\ \leq F _ {M} \left(\mu_ {2}, z _ {2}\right) + L _ {f} d _ {Z} \left(z _ {1}, z _ {2}\right) - L _ {f} d _ {\mathrm {K R}} \left(\hat {\mu} [ \vec {x} _ {2} ], \mu_ {2}\right) + L _ {f} d _ {\mathrm {K R}} \left(\mu_ {1}, \mu_ {2}\right) \\ + L _ {f} d _ {\mathrm {K R}} \left(\hat {\mu} [ \vec {x} _ {2} ], \mu_ {2}\right) \\ = F _ {M} \left(\mu_ {2}, z _ {2}\right) + L _ {f} \left(d _ {\mathrm {K R}} \left(\mu_ {1}, \mu_ {2}\right) + d _ {Z} \left(z _ {1}, z _ {2}\right)\right), \\ \end{array}
$$
where we used the definition of  $F_{M}$  in the first inequality, the Lipschitz continuity of  $f_{M}$  (w.r.t. the second argument) for the second inequality, and then the fact that  $\vec{x}_2$  attains the infimum in the definition of  $F_{M}(\mu_{2}, z_{2})$  and the triangle inequality for  $d_{\mathrm{KR}}$ . Interchanging the roles of  $\mu_{1}, z_{1}$  and  $\mu_{2}, z_{2}$  then establishes the claim.
Step 3 There exists  $B_F \in \mathbb{R}_{\geq 0}$  such that for all  $M \in \mathbb{N}_+$ ,  $\mu \in \mathcal{P}(X)$  and  $z \in Z$  we have  $|F_M(\mu, z)| \leq B_F + h(z)$ : Let  $D_{\mathcal{P}(X)}$  be the diameter of  $\mathcal{P}(X)$  (which is finite since  $\mathcal{P}(X)$  is compact), then for all  $M \in \mathbb{N}_+$  and  $\vec{x} \in X^M$ ,  $z \in Z$ ,  $\mu \in \mathcal{P}(X)$  we have
$$
- \left(B _ {f} + L _ {f} D _ {\mathcal {P} (X)} + b (z)\right) \leq f _ {M} (\vec {x}, z) + L _ {f} d _ {\mathrm {K R}} (\hat {\mu} [ \vec {x} ], \mu) \leq B _ {f} + L _ {f} D _ {\mathcal {P} (X)} + b (z),
$$
therefore  $|F_M(\mu, z)| \leq B_f + L_f D_{\mathcal{P}(X)} + b(z)$ , showing the claim with  $B_F = B_f + L_f D_{\mathcal{P}(X)}$ .
Step 4 Summarizing,  $(F_M)_M$  is a sequence of  $L_{f}$ -Lipschitz continuous and hence equicontinuous functions such that for all  $\mu \in \mathcal{P}(X)$  and  $z \in Z$ , the set  $\{F_M(\mu, z) \mid M \in \mathbb{N}_+\}$  is relatively compact (since it is a bounded subset of  $\mathbb{R}$ ). We can now use a variant of the Arzela-Ascoli theorem, cf. [30,
Corollary III.3.3]. From the assumption on  $Z$ , we can find a sequence  $(V_{n})_{n}$  of open subsets of  $Z$  such that all  $\bar{V}_n$  are compact,  $\bar{V}_n \subseteq V_{n + 1}$  and we have  $\bigcup_{n} V_{n} = Z$ . Then  $(F_M|_{\bar{V}_n})_M$  is a sequence of functions that fulfills the conditions of the Arzela-Ascoli theorem (since  $\mathcal{P}(X) \times K_n$  is compact), so there exists a subsequence  $(F_{M_\ell^{(n)}}|_{\bar{V}_n})_\ell$  that converges uniformly to a continuous function on  $\mathcal{P}(X) \times \bar{V}_n$ . Denote the diagonal subsequence of all these subsequences by  $(F_{M_\ell})_\ell$ , then there exists a continuous  $f: \mathcal{P}(X) \times Z \to \mathbb{R}$  such that  $(F_{M_\ell})_\ell$  converges uniformly on compact subsets to  $f$ . Since  $\mathcal{P}(X)$  is compact, this means that for all compact  $K \subseteq Z$
$$
\lim_{\ell}\sup_{\substack{\mu \in \mathcal{P}(X)\\ z\in K}}|F_{M_{\ell}}(\mu ,z) - f(\mu ,z)| = 0.
$$
This also implies that for all  $\mu \in \mathcal{P}(X)$  and  $z\in Z$  we have  $|f(\mu ,z)|\leq B_F + b(z)$
Furthermore,  $f$  is also  $L_{f}$ -Lipschitz continuous: Let  $\mu_{i} \in \mathcal{P}(X), z_{i} \in Z, i = 1,2$ , and  $\epsilon > 0$  be arbitrary. Let  $K \subseteq Z$  be compact with  $z_{1}, z_{2} \in K$  and choose  $\ell \in \mathbb{N}_{+}$  such that
$$
\sup_{\substack{\mu \in \mathcal{P}(X)\\ z\in K}}|F_{M_{\ell}}(\mu ,z) - f(\mu ,z)|\leq \frac{\epsilon}{2}.
$$
$$
\begin{array}{l} \left| f \left(\mu_ {1}, z _ {1}\right) - f \left(\mu_ {2}, z _ {2}\right) \right| \leq \left| f \left(\mu_ {1}, z _ {1}\right) - F _ {M _ {\ell}} \left(\mu_ {1}, z _ {1}\right) \right| + \left| F _ {M _ {\ell}} \left(\mu_ {1}, z _ {1}\right) - F _ {M _ {\ell}} \left(\mu_ {2}, z _ {2}\right) \right| \\ + \left| F _ {M _ {\ell}} \left(\mu_ {2}, z _ {2}\right) - f \left(\mu_ {2}, z _ {2}\right) \right| \\ \leq L _ {f} \left(d _ {\mathrm {K R}} \left(\mu_ {1}, \mu_ {2}\right) + d _ {Z} \left(z _ {1}, z _ {2}\right)\right) + \epsilon , \\ \end{array}
$$
and since  $\epsilon > 0$  was arbitrary, the claim follows.
Step 5 For  $\ell \in \mathbb{N}_{+}$  and  $\vec{x} \in X^{M_{\ell}}$ ,  $z \in Z$  we have
$$
\left| f _ {M _ {\ell}} (\vec {x}, z) - f \left(\hat {\mu} [ \vec {x} ], z\right) \right| = \left| F _ {M _ {\ell}} \left(\hat {\mu} [ \vec {x} ], z\right) - f \left(\hat {\mu} [ \vec {x} ], z\right) \right|
$$
since  $F_{M_{\ell}}$  extends  $f_{M_{\ell}}$ , and hence
$$
\sup_{\substack{\vec{x}\in X^{M_{\ell}}\\ z\in K}}|f_{M_{\ell}}(\vec{x},z) - f(\hat{\mu} [\vec{x}],z)|\to 0.
$$
Next, we provide the proofs for the  $\Gamma$ -lim inf and  $\Gamma$ -lim sup results.
Proof. of Lemma 2.4 Assume the statement is not true, i.e.,  $\| f \|_k > \lim \inf_{M \to \infty} \| f_M \|_M$ . This means that there exists a subsequence  $M_\ell$  and  $C \in \mathbb{R}_{\geq 0}$  such that  $\| f \|_k > \lim_\ell \| f_{M_\ell} \|_{M_\ell} = C$ . Note that this implies that  $\| f \|_k > 0$ .
Let  $\epsilon_1, \epsilon_2 > 0$  and  $\alpha > 1$ ,  $\beta \in (0,1)$  be arbitrary. From Theorem B.1, there exists  $(\vec{\mu}, \vec{\alpha}) \in \mathcal{P}(X)^N \times \mathbb{R}^N$  such that
$$
\mathcal {D} (\vec {\mu}, \vec {\alpha}, f, k) + \epsilon_ {1} \geq \| f \| _ {k},
$$
and w.l.o.g. we can assume that  $\epsilon_1 > 0$  is small enough so that  $\mathcal{D}(\vec{\mu},\vec{\alpha},f,k) > 0$ . The latter implies that  $\mathcal{E}(\vec{\mu},\vec{\alpha},f)$ ,  $\mathcal{W}(\vec{\mu},\vec{\alpha},k) > 0$ , so defining
$$
\begin{array}{l} \epsilon_ {\alpha} = \frac {\alpha - 1}{\alpha} \mathcal {E} (\vec {\mu}, \vec {\alpha}, f) \\ \epsilon_ {\beta} = (1 / \beta - 1) \mathcal {W} (\vec {\mu}, \vec {\alpha}, k) \\ \end{array}
$$
we get  $\epsilon_{\alpha},\epsilon_{\beta} > 0$  . For each  $n = 1,\ldots ,N$  , choose  $\vec{x}_n^{[M]}\in X^M$  such that  $\vec{x}_n^{[M]}\xrightarrow{d_{\mathrm{KR}}}\mu_n$  for  $M\to \infty$  Choose now  $L_{1}\in \mathbb{N}$  such that for all  $\ell \geq L_1$  we get
$$
\left| \mathcal {E} \left(\vec {X} ^ {[ M _ {\ell} ]}, \vec {\alpha}, f _ {M _ {\ell}}\right) - \mathcal {E} (\vec {\mu}, \vec {\alpha}, f) \right| \leq \epsilon_ {\alpha}
$$
$$
\left| \mathcal {W} \left(\vec {X} ^ {[ M _ {\ell} ]}, \vec {\alpha}, k _ {M _ {\ell}}\right) - \mathcal {W} (\vec {\mu}, \vec {\alpha}, k) \right| \leq \epsilon_ {\beta}.
$$
(cf. also the proof of Theorem 2.3) and  $\mathcal{W}(\vec{X}^{[M_{\varepsilon}]},\vec{\alpha},k^{[M_{\varepsilon}]}) > 0$  . We then get
$$
\begin{array}{l} \mathcal {E} (\vec {\mu}, \vec {\alpha}, f) \leq \alpha \mathcal {E} (\vec {X} ^ {[ M _ {\ell} ]}, \vec {\alpha}, f _ {M _ {\ell}}) \\ \mathcal {W} (\vec {\mu}, \vec {\alpha}, k) \geq \beta \mathcal {W} (\vec {X} ^ {[ M _ {\ell} ]}, \vec {\alpha}, k ^ {[ M _ {\ell} ]}), \\ \end{array}
$$
so altogether
$$
\frac {\mathcal {E} (\vec {\mu} , \vec {\alpha} , f)}{\mathcal {W} (\vec {\mu} , \vec {\alpha} , k)} \leq \frac {\alpha \mathcal {E} (\vec {X} ^ {[ M _ {\ell} ]} , \vec {\alpha} , f _ {M _ {\ell}})}{\beta \mathcal {W} (\vec {X} ^ {[ M _ {\ell} ]} , \vec {\alpha} , k ^ {[ M _ {\ell} ]})}.
$$
Using Theorem B.1 again leads to
$$
\frac {\alpha \mathcal {E} (\vec {X} ^ {[ M _ {\ell} ]} , \vec {\alpha} , f _ {M})}{\beta \mathcal {W} (\vec {X} ^ {[ M _ {\ell} ]} , \vec {\alpha} , k ^ {[ M _ {\ell} ]})} = \mathcal {D} (\vec {X} ^ {[ M _ {\ell} ]}, \vec {\alpha}, f _ {M _ {\ell}}, k ^ {[ M _ {\ell} ]}) \leq \| f _ {M _ {\ell}} \| _ {M _ {\ell}}.
$$
Finally, let  $L_{2}$  such that for all  $\ell \geq L_{2}$  we have  $\| f_{M_{\ell}} \|_{M_{\ell}} \leq C + \epsilon_{2}$ . For  $\ell \geq L_{1}, L_{2}$  we then get
$$
\begin{array}{l} C <   \| f \| _ {k} \leq \mathcal {D} (\vec {\mu}, \vec {\alpha}, f, k) + \epsilon_ {1} \\ = \frac {\mathcal {E} (\vec {\mu} , \vec {\alpha} , f)}{\mathcal {W} (\vec {\mu} , \vec {\alpha} , k)} + \epsilon_ {1} \\ \leq \frac {\alpha \mathcal {E} (\vec {X} ^ {[ M _ {\ell} ]} , \vec {\alpha} , f _ {M _ {\ell}})}{\beta \mathcal {W} (\vec {X} ^ {[ M _ {\ell} ]} , \vec {\alpha} , k ^ {[ M _ {\ell} ]})} + \epsilon_ {1} \\ \leq \frac {\alpha}{\beta} \| f _ {M _ {\ell}} \| _ {M _ {\ell}} + \epsilon_ {1} \\ \leq \frac {\alpha}{\beta} C + \frac {\alpha}{\beta} \epsilon_ {2} + \epsilon_ {1}. \\ \end{array}
$$
Since  $\epsilon_1, \epsilon_2 > 0$  and  $\alpha > 1$ ,  $\beta \in (0,1)$  were arbitrary, this implies that
$$
C <   \| f \| _ {k} \leq C,
$$
Proof. of Lemma 2.5 Let  $f \in H_{k}$  be arbitrary and choose  $(\epsilon_{n})_{n} \subseteq \mathbb{R}_{>0}$  with  $\epsilon_{n} \searrow 0$ .
Step 1 For each  $n\in \mathbb{N}$  choose
$$
f _ {n} ^ {\text {p r e}} = \sum_ {\ell = 1} ^ {L _ {n}} \alpha_ {\ell} ^ {(n)} k (\cdot , \mu_ {\ell} ^ {(n)}) \in H _ {k} ^ {\text {p r e}},
$$
where  $\alpha_{1}^{(n)},\ldots ,\alpha_{L_{n}}^{(n)}\in \mathbb{R}$  and  $\mu_1^{(n)},\dots,\mu_{L_n}^{(n)}\in \mathcal{P}(X)$ , with
$$
\left\| f - f _ {n} ^ {\text {p r e}} \right\| _ {k} \leq \frac {\epsilon_ {n}}{3 \sqrt {C _ {k}}}
$$
and  $\| f_n^{\mathrm{pre}}\| _k\leq \| f\| _k$  . To see that such a sequence of functions exists, choose some sequence  $(\bar{f}_n)_n\in H_k^{\mathrm{pre}}$  with  $\bar{f}_n = \sum_{\ell = 1}^{\bar{L}_n}\bar{\alpha}_{\ell}^{(n)}k(\cdot ,\bar{\mu}_{\ell}^{(n)})$  , where  $\bar{\alpha}_{\ell}^{(n)}\in \mathbb{R},\bar{\mu}_{\ell}^{(n)}\in \mathcal{P}(X)$  , with  $\bar{f}_n\xrightarrow{\|\cdot\|_k}f$  (exists since  $H_{k}^{\mathrm{pre}}$  is dense in  $H_{k}$  ). Define now for  $n\in \mathbb{N}$
$$
\bar {H} _ {n} = \operatorname {s p a n} \left\{k \left(\cdot , \bar {\mu} _ {\ell} ^ {(m)}\right) \mid m = 1, \dots , n, \ell = 1, \dots , \bar {L} _ {m} \right\}
$$
and  $\hat{f}_n = P_{\bar{H}_n}f$ , where  $P_{\bar{H}_n}$  is the orthogonal projection onto  $\bar{H}_n$ . Then  $\bar{H}_n \subseteq H_k^{\mathrm{pre}}$ ,  $\| \hat{f}_n \|_k = \| P_{\bar{H}_n}f \|_k \leq \| f \|_k$  and  $\| f - \hat{f}_n \|_k \leq \| f - \bar{f}_n \|_k \to 0$  (since  $\hat{f}_n = P_{\bar{H}_n}f$  is the orthogonal projection of  $f$  onto  $\bar{H}_n$  and  $\bar{f}_n \in \bar{H}_n$ ), hence  $\hat{f}_n \xrightarrow{\|\cdot\|_k} f$ . We can now choose  $(f_n^{\mathrm{pre}})_n$  as a subsequence of  $(\hat{f}_n)_n$ .
Next, for all  $n \in \mathbb{N}$  and  $\ell = 1, \ldots, L_n$  choose  $\vec{x}_M^{(n,\ell)} \in X^M$  with  $\hat{\mu}[\vec{x}_M^{(n,\ell)}] \xrightarrow{d_{\mathrm{KR}}}\mu_\ell^{(n)}$  for  $M \to \infty$ . Furthermore, for all  $n \in \mathbb{N}$  choose  $M_n \in \mathbb{N}$  such that for all  $M \geq M_n$  and  $\ell = 1, \ldots, L_n$  we have
$$
d _ {\mathrm {K R}} (\hat {\mu} [ \vec {x} _ {M} ^ {(n, \ell)} ], \mu_ {\ell} ^ {(n)}) \leq \min \left\{\frac {\epsilon_ {n}}{3 (1 + L _ {k} \sum_ {\ell^ {\prime} = 1} ^ {L _ {n}} | \alpha_ {\ell^ {\prime}} ^ {(n)} |)}, \frac {\epsilon_ {n} ^ {2}}{2 (1 + 2 L _ {k} \sum_ {i , j = 1} ^ {L _ {n}} | \alpha_ {i} ^ {(n)} | | \alpha_ {j} ^ {(n)} |)} \right\}
$$
$$
\sup  _ {\vec {x}, \vec {x} ^ {\prime} \in X ^ {M}} | k _ {M} (\vec {x}, \vec {x} ^ {\prime}) - k (\hat {\mu} [ \vec {x} ], \hat {\mu} [ \vec {x} ^ {\prime} ]) | \leq \min  \left\{\frac {\epsilon_ {n}}{3 \left(1 + \sum_ {\ell^ {\prime} = 1} ^ {L _ {n}} | \alpha_ {\ell^ {\prime}} ^ {(n)} |\right)}, \frac {\epsilon_ {n} ^ {2}}{2 \left(1 + \sum_ {i , j = 1} ^ {L _ {n}} | \alpha_ {i} ^ {(n)} | | \alpha_ {j} ^ {(n)} |\right)} \right\}.
$$
W.l.o.g. we can assume that  $(M_n)_n$  is strictly increasing. For  $M \in \mathbb{N}$ , let  $n(M)$  be the largest integer such that  $M_{n(M)} \leq M$  and define
$$
\hat {f} _ {M} ^ {\text {p r e}} = \sum_ {\ell = 1} ^ {L _ {n (M)}} \alpha_ {\ell} ^ {(n (M))} k (\cdot , \hat {\mu} [ \vec {x} _ {M} ^ {(n (M), \ell)} ]) \in H _ {k} ^ {\text {p r e}}
$$
$$
f _ {M} = \sum_ {\ell = 1} ^ {L _ {n (M)}} \alpha_ {\ell} ^ {(n (M))} k _ {M} (\cdot , \vec {x} _ {M} ^ {(n (M), \ell)}) \in H _ {M} ^ {\mathrm {p r e}}.
$$
Step 2 We now show that  $f_{M} \xrightarrow{\mathcal{P}_{1}} f$ . For this, let  $\epsilon > 0$  be arbitrary and  $n_{\epsilon} \in \mathbb{N}$  such that  $\epsilon_{n} \leq \epsilon$ . Let now  $M \geq M_{n_{\epsilon}}$  (note that this implies that  $n(M) \geq n_{\epsilon}$  and hence  $\epsilon_{n(M)} \leq \epsilon_{n}$ ) and  $\vec{x} \in X^{M}$ , then we have
$$
|f(\hat{\mu} [\vec{x} ])- f_{M}(\vec{x})|\leq \underbrace{|f(\hat{\mu}[\vec{x} ]) - f_{n(M)}(\hat{\mu}[\vec{x}])|}_{= I} + \underbrace{|f_{n(M)}(\hat{\mu}[\vec{x} ]) - \hat{f}_{M}^{\text{pre}}(\hat{\mu}[\vec{x}])|}_{= II} + \underbrace{|\hat{f}_{M}^{\text{pre}}(\hat{\mu}[\vec{x} ]) - f_{M}(\vec{x})|}_{= III}
$$
We continue with
$$
\begin{array}{l} I = \left| f (\hat {\mu} [ \vec {x} ]) - f _ {n (M)} (\hat {\mu} [ \vec {x} ]) \right| \\ = | \langle f - f _ {n (M)}, k (\cdot , \hat {\mu} [ \vec {x} ]) \rangle_ {k} | \\ \leq \| f - f _ {n (M)} \| _ {k} \| k (\cdot , \hat {\mu} [ \vec {x} ]) \| _ {k} \\ = \| f - f _ {n (M)} \| _ {k} \sqrt {k (\hat {\mu} [ \vec {x} ] , \hat {\mu} [ \vec {x} ])} \\ \leq \frac {\epsilon_ {n (M)}}{3 \sqrt {C _ {k}}} \sqrt {C _ {k}} \\ \end{array}
$$
where we first used the reproducing property of  $k$ , then Cauchy-Schwarz, again the reproducing property of  $k$ , and finally the choice  $f_{n(M)}$  and the boundedness of  $k$ .
$$
\begin{array}{l} I I = \left| f _ {n (M)} (\hat {\mu} [ \vec {x} ]) - \hat {f} _ {M} ^ {\mathrm {p r e}} (\hat {\mu} [ \vec {x} ]) \right| \\ = \left| \sum_ {\ell = 1} ^ {L _ {n (M)}} \alpha_ {\ell} ^ {(n (M))} k (\cdot , \mu_ {\ell} ^ {(n (M))}) - \sum_ {\ell = 1} ^ {L _ {n (M)}} \alpha_ {\ell} ^ {(n (M))} k (\cdot , \hat {\mu} [ \vec {x} _ {M} ^ {(n (M), \ell)} ]) \right| \\ \leq \sum_ {\ell = 1} ^ {L _ {n (M)}} \left| \alpha_ {\ell} ^ {(n (M))} \right| | k (\cdot , \mu_ {\ell} ^ {(n (M))}) - k (\cdot , \hat {\mu} [ \vec {x} _ {M} ^ {(n (M), \ell)} ]) | \\ \leq L _ {k} \sum_ {\ell = 1} ^ {L _ {n (M)}} \left| \alpha_ {\ell} ^ {(n (M))} \right| d _ {\mathrm {K R}} \left(\hat {\mu} \left[ \vec {x} _ {M} ^ {(n (M), \ell)} \right], \mu_ {\ell} ^ {(n (M))}\right) \\ \leq \frac {\epsilon_ {n (M)}}{3}, \\ \end{array}
$$
where we used the triangle inequality, the Lipschitz continuity of  $k$ , and then the choice of the sequence  $(M_n)_n$ .
$$
\begin{array}{l} I I I = \left| \hat {f} _ {M} ^ {\text {p r e}} (\hat {\mu} [ \vec {x} ]) - f _ {M} (\vec {x}) \right| \\ = \left| \sum_ {\ell = 1} ^ {L _ {n (M)}} \alpha_ {\ell} ^ {(n (M))} k (\cdot , \hat {\mu} [ \vec {x} _ {M} ^ {(n (M), \ell)} ]) - \sum_ {\ell = 1} ^ {L _ {n (M)}} \alpha_ {\ell} ^ {(n (M))} k _ {M} (\cdot , \vec {x} _ {M} ^ {(n (M), \ell)}) \right| \\ \leq \sum_ {\ell = 1} ^ {L _ {n (M)}} \left| \alpha_ {\ell} ^ {(n (M))} \right| | k (\cdot , \hat {\mu} [ \vec {x} _ {M} ^ {(n (M), \ell)} ]) - k _ {M} (\cdot , \vec {x} _ {M} ^ {(n (M), \ell)}) | \\ \leq \frac {\epsilon_ {n (M)}}{3}, \\ \end{array}
$$
where the triangle inequality has been used in the first step and then again the choice of the sequence  $(M_n)_n$ .
$$
\begin{array}{l} \left| f \left(\hat {\mu} [ \vec {x} ]\right) - f _ {M} (\vec {x}) \right| \leq I + I I + I I I \\ \leq \frac {\epsilon_ {n (M)}}{3} + \frac {\epsilon_ {n (M)}}{3} + \frac {\epsilon_ {n (M)}}{3} \\ \leq \epsilon , \\ \end{array}
$$
establishing  $f_{M}\xrightarrow{\mathcal{P}_{1}}f$
Step 3 We now show  $\lim \sup_{M\to \infty}\| f_M\| _M\leq \| f\| _k$ . Let  $\epsilon >0$  be arbitrary and  $n_{\epsilon}\in \mathbb{N}$  such that  $\epsilon_{n}\leq \epsilon$  and let  $M\geq M_{n_{\epsilon}}$ . We have
$$
\begin{array}{l} \| f _ {M} \| _ {M} ^ {2} = \sum_ {\ell , \ell^ {\prime} = 1} ^ {L _ {n (M)}} \alpha_ {\ell} ^ {(n (M))} \alpha_ {\ell^ {\prime}} ^ {(n (M))} k _ {M} (\vec {x} _ {M} ^ {(n (M), \ell^ {\prime})}, \vec {x} _ {M} ^ {(n (M), \ell^ {\prime})}) \\ \leq \sum_ {\ell , \ell^ {\prime} = 1} ^ {L _ {n (M)}} \alpha_ {\ell} ^ {(n (M))} \alpha_ {\ell^ {\prime}} ^ {(n (M))} k \left(\mu_ {\ell^ {\prime}} ^ {(n (M))}, \mu_ {\ell} ^ {(n (M))}\right) + | R _ {1} | + | R _ {2} | \\ = \| f _ {n (M)} ^ {\text {p r e}} \| _ {k} ^ {2} + R _ {1} + R _ {2} \\ \leq \| f \| _ {k} ^ {2} + R _ {1} + R _ {2}. \\ \end{array}
$$
with remainder terms
$$
\begin{array}{l} R _ {1} = \sum_ {\ell , \ell^ {\prime} = 1} ^ {L _ {n (M)}} \alpha_ {\ell} ^ {(n (M))} \alpha_ {\ell^ {\prime}} ^ {(n (M))} k _ {M} (\vec {x} _ {M} ^ {(n (M), \ell^ {\prime})}, \vec {x} _ {M} ^ {(n (M), \ell^ {\prime})}) - \sum_ {\ell , \ell^ {\prime} = 1} ^ {L _ {n (M)}} \alpha_ {\ell} ^ {(n (M))} \alpha_ {\ell^ {\prime}} ^ {(n (M))} k (\hat {\mu} [ \vec {x} _ {M} ^ {(n (M), \ell^ {\prime})} ], \hat {\mu} [ \vec {x} _ {M} ^ {(n (M), \ell^ {\prime})} ]) \\ R _ {2} = \sum_ {\ell , \ell^ {\prime} = 1} ^ {L _ {n (M)}} \alpha_ {\ell} ^ {(n (M))} \alpha_ {\ell^ {\prime}} ^ {(n (M))} k (\hat {\mu} [ \vec {x} _ {M} ^ {(n (M), \ell^ {\prime})} ], \hat {\mu} [ \vec {x} _ {M} ^ {(n (M), \ell^ {\prime})} ]) - \sum_ {\ell , \ell^ {\prime} = 1} ^ {L _ {n (M)}} \alpha_ {\ell} ^ {(n (M))} \alpha_ {\ell^ {\prime}} ^ {(n (M))} k (\mu_ {\ell^ {\prime}} ^ {(n (M))}, \mu_ {\ell} ^ {(n (M))}) \\ \end{array}
$$
We now bound these terms, so that
$$
\begin{array}{l} R _ {1} = \left| \sum_ {\ell , \ell^ {\prime} = 1} ^ {L _ {n (M)}} \alpha_ {\ell} ^ {(n (M))} \alpha_ {\ell^ {\prime}} ^ {(n (M))} k _ {M} (\vec {x} _ {M} ^ {(n (M), \ell^ {\prime})}, \vec {x} _ {M} ^ {(n (M), \ell^ {\prime})}) - \sum_ {\ell , \ell^ {\prime} = 1} ^ {L _ {n (M)}} \alpha_ {\ell} ^ {(n (M))} \alpha_ {\ell^ {\prime}} ^ {(n (M))} k (\hat {\mu} [ \vec {x} _ {M} ^ {(n (M), \ell^ {\prime})} ], \hat {\mu} [ \vec {x} _ {M} ^ {(n (M), \ell^ {\prime})} ]) \right| \\ \leq \sum_ {\ell , \ell^ {\prime} = 1} ^ {L _ {n (M)}} | \alpha_ {\ell} ^ {(n (M))} | | \alpha_ {\ell^ {\prime}} ^ {(n (M))} | | k _ {M} (\vec {x} _ {M} ^ {(n (M), \ell^ {\prime})}, \vec {x} _ {M} ^ {(n (M), \ell^ {\prime})}) - k (\hat {\mu} [ \vec {x} _ {M} ^ {(n (M), \ell^ {\prime})} ], \hat {\mu} [ \vec {x} _ {M} ^ {(n (M), \ell^ {\prime})} ]) | \\ \leq \frac {\epsilon_ {n (M)} ^ {2}}{2}, \\ \end{array}
$$