SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \mathcal {L} _ {P} ^ {u ^ {*}, \mathrm {a b s t}} (\hat {h}) \leq \mathcal {L} _ {P} ^ {u ^ {*}, \mathrm {a b s t}} (h ^ {*}) + P _ {\mathcal {X}} \left(\left\{x \in \mathcal {X}: h ^ {*} (x) \neq \hat {h} (x) \right\}\right) \\ \leq \mathcal {L} _ {P} ^ {u ^ {*}} \left(h ^ {*}\right) + P _ {\mathcal {X}} \left(\left\{x \in \mathcal {X}: \hat {h} (x) = \star \right\}\right) \\ 0 + P _ {\mathcal {X}} \left(\operatorname {D I S} \left(B \left(h ^ {*}, \frac {\epsilon}{\theta (\mathcal {H})}\right)\right) = \epsilon . \right. \\ \end{array}
$$
D.7 Proof of Theorem 6
Proof. Let  $m \geq m_{\mathcal{U},\mathcal{H},\text{cover}}(\epsilon /2,\epsilon /2,\delta)$  and  $S \sim P^m$ . And let  $u^{*} \in \mathcal{U}$  the ground-truth perturbation set. We define the abstention learner as:
$$
\mathcal {A} (S) (x) = \left\{ \begin{array}{l l} 1 & \text {i f f o r a l l h \in \mathcal {A} _ {c o v e r} (S) : h (x) = 1} \\ 0 & \text {i f f o r a l l h \in \mathcal {A} _ {c o v e r} (S) : h (x) = 0} \\ \star & \text {o t h e r w i s e} \end{array} \right.
$$
Then with probability  $1 - \delta$ ,  $P_{\mathcal{X}}(\mathrm{DIS}(\mathcal{A}_{\mathrm{cover}}(S))) \leq \frac{\epsilon}{2}$  and  $\exists h \in \mathcal{A}_{\mathrm{cover}}(S)$  with
$$
\mathcal {L} _ {P} ^ {u ^ {*}} (h) \leq \min  _ {h ^ {\prime} \in \mathcal {H}} \mathcal {L} _ {P} ^ {u ^ {*}} \left(h ^ {\prime}\right) + \epsilon / 2.
$$
Since for every  $x\in \mathcal{X}$  either  $A(S)(x) = h(x)$  or  $x\in \mathcal{A}_{\text{cover}}(S)$ . Thus,
$$
\mathcal {L} _ {P} ^ {u ^ {*}, \mathrm {a b s t}} (A (S)) \leq \mathcal {L} _ {P} ^ {u ^ {*}, \mathrm {a b s t}} (h) + P _ {\mathcal {X}} (\operatorname {D I S} (\mathcal {A} _ {c o v e r} (S))) \leq \min  _ {h ^ {\prime} \in \mathcal {H}} \mathcal {L} _ {P} ^ {u ^ {*}, \mathrm {a b s t}} (h ^ {\prime}) + \epsilon .
$$