SlowGuess/NeurIPS_2023_OCR huggingface dataset:
where the inequality is by event  $\mathcal{E}_2$  and because  $A' \in \widetilde{\mathcal{A}}_{p,d}$  and  $\mathbb{P}(X \in A') \geq \mathbb{P}(X \in A) \geq \bar{t}(\delta/2) \geq \bar{t}_1(\delta/2)$ . Note that
$$
\mathbb {P} \left(X \in A ^ {\prime} \backslash A\right) \leq \frac {2 \bar {\theta} d}{n} \leq \bar {t} _ {2} (\delta / 2) \leq \mathbb {P} (X \in A) \tag {54}
$$
where the first inequality is by Lemma A.5 (2); the second inequality is by the definition of  $\hat{t}_2(\delta / 2)$  in (22); the third inequality is because  $A \in \mathcal{A}_{p,d} \setminus \mathcal{A}_{p,d}(\bar{t}(\delta / 2))$ . As a result of (53) and (54), we have
$$
T _ {2} \leq \sqrt {\frac {\mathbb {P} \left(X \in A ^ {\prime} \backslash A\right) + \mathbb {P} \left(X \in A\right)}{\mathbb {P} (X \in A)}} 2 V \sqrt {\bar {t} _ {1} (\delta / 2)} \leq 2 \sqrt {2} V \sqrt {\bar {t} _ {1} (\delta / 2)} \tag {55}
$$
To bound  $T_{3}$ , note that
$$
\begin{array}{l} T _ {3} = \frac {1}{\sqrt {\mathbb {P} (X \in A)}} \left| \mathbb {E} \left(z _ {1} 1 _ {\{x _ {1} \in A ^ {\prime} \setminus A \}}\right) \right| \leq \frac {V}{\sqrt {\mathbb {P} (X \in A)}} \mathbb {P} \left(X \in A ^ {\prime} \setminus A\right) \tag {56} \\ \leq V \sqrt {\mathbb {P} \left(X \in A ^ {\prime} \backslash A\right)} \leq V \sqrt {2 \bar {\theta} d / n} \leq V \sqrt {\bar {t} _ {2} (\delta / 2)} \\ \end{array}
$$
The proof is complete by combining inequalities (51), (52), (55) and (56), and note that
$$
2 V \sqrt {\bar {t} _ {2} (\delta / 2)} + 2 \sqrt {2} V \sqrt {\bar {t} _ {1} (\delta / 2)} \leq 5 V \sqrt {\bar {t} (\delta / 2)}.
$$
Now we are ready to wrap up the proof of Lemma A.1.
Completing the proof of Lemma A.1
Define events  $\mathcal{E}_1$  and  $\mathcal{E}_2$ :
$$
\mathcal {E} _ {1} := \left\{\sup  _ {A \in \mathcal {A} _ {p, d} \backslash \mathcal {A} _ {p, d} (\bar {t} (\delta / 8))} \frac {1}{\sqrt {\mathbb {P} (X \in A)}} \left| \frac {1}{n} \sum_ {i = 1} ^ {n} 1 _ {\{x _ {i} \in A \}} - \mathbb {P} (X \in A) \right| \leq 5 \sqrt {\bar {t} (\delta / 8)} \right\}
$$
$$
\mathcal {E} _ {2} := \left\{\sup  _ {A \in \mathcal {A} _ {p, d} \backslash \mathcal {A} _ {p, d} (\bar {t} (\delta / 8))} \frac {1}{\sqrt {\mathbb {P} (X \in A)}} \Big | \frac {1}{n} \sum_ {i = 1} ^ {n} y _ {i} 1 _ {\{x _ {i} \in A \}} - \mathbb {E} (y _ {1} 1 _ {\{x _ {1} \in A \}}) \Big | \leq 5 U \sqrt {\bar {t} (\delta / 8)} \right\}
$$
Then by Lemma A.9 with  $z_{i} = y_{i}$  and  $z_{i} = 1$  respectively, we know that  $\mathbb{P}(\mathcal{E}_i)\geq 1 - \delta /4$  for all  $i = 1,2$ . So we know  $\mathbb{P}(\cap_{i = 1}^{2}\mathcal{E}_{i})\geq 1 - \delta$ . Below we prove that inequality (23) is true when  $\cap_{i = 1}^{2}\mathcal{E}_{i}$  hold.
Define  $a\coloneqq 100\bar{t} (\delta /8)$  . Then it holds
$$
\sup  _ {A \in \mathcal {A} _ {p, d} (a)} \sqrt {\mathbb {P} (X \in A)} \left| \mathbb {E} \left(f ^ {*} (X) \mid X \in A\right) - \bar {y} _ {\mathcal {I} _ {A}} \right| \leq 2 U \sqrt {a} = 2 0 U \sqrt {\bar {t} (\delta / 8)} \tag {57}
$$
On the other hand, for any  $A \in \mathcal{A}_{p,d} \setminus \mathcal{A}_{p,d}(a)$ , by event  $\mathcal{E}_1$ , we have
$$
\frac {1}{n} \sum_ {i = 1} ^ {n} 1 _ {\{x _ {i} \in A \}} \geq \mathbb {P} (X \in A) - 5 \sqrt {\bar {t} (\delta / 8)} \sqrt {\mathbb {P} (X \in A)} \geq \frac {1}{2} \mathbb {P} (X \in A) \tag {58}
$$
where the second inequality is because  $\mathbb{P}(X\in A)\geq a = 100\bar{t} (\delta /8)$ . Therefore we know  $\sum_{i = 1}^{n}1_{\{x_i\in A\}} > 0$  and we can write
$$
\begin{array}{l} \mathbb {E} (f ^ {*} (X) | X \in A) - \bar {y} _ {\mathcal {I} _ {A}} = \frac {\mathbb {E} (y _ {1} 1 _ {\{x _ {1} \in A \}})}{\mathbb {P} (X \in A)} - \frac {\frac {1}{n} \sum_ {i = 1} ^ {n} y _ {i} 1 _ {\{x _ {i} \in A \}}}{\frac {1}{n} \sum_ {i = 1} ^ {n} 1 _ {\{x _ {i} \in A \}}} \\ = \frac {1}{\mathbb {P} (X \in A)} \left(\mathbb {E} \left(y _ {1} 1 _ {\{x _ {1} \in A \}}\right) - \frac {1}{n} \sum_ {i = 1} ^ {n} y _ {i} 1 _ {\{x _ {i} \in A \}}\right) \tag {59} \\ + \frac {\sum_ {i = 1} ^ {n} y _ {i} \mathrm {l} _ {\{x _ {i} \in A \}}}{\sum_ {i = 1} ^ {n} 1 _ {\{x _ {i} \in A \}} \mathbb {P} (X \in A)} \left(\frac {1}{n} \sum_ {i = 1} ^ {n} 1 _ {\{x _ {i} \in A \}} - \mathbb {P} (X \in A)\right) \\ := H _ {1} (A) + H _ {2} (A) \\ \end{array}
$$
By event  $\mathcal{E}_2$ , and note that  $a \geq \bar{t}(\delta/8)$ , we have
$$
\sup  _ {A \in \mathcal {A} _ {p, d} \backslash \mathcal {A} _ {p, d} (a)} \sqrt {\mathbb {P} (X \in A)} | H _ {1} (A) | \leq 5 U \sqrt {\bar {t} (\delta / 8)} \tag {60}
$$
By event  $\mathcal{E}_1$ , and note that  $a \geq \bar{t}(\delta/8)$ , we have
$$
\sup  _ {A \in \mathcal {A} _ {p, d} \backslash \mathcal {A} _ {p, d} (a)} \sqrt {\mathbb {P} (X \in A)} | H _ {2} (A) | \leq 5 U \sqrt {t (\delta / 8)} \tag {61}
$$
Combining (59), (60) and (61), we have
$$
\sup  _ {A \in \mathcal {A} _ {p, d} \backslash \mathcal {A} _ {p, d} (a)} \sqrt {\mathbb {P} (X \in A)} \left| \mathbb {E} (f ^ {*} (X) | X \in A) - \bar {y} _ {\mathcal {I} _ {A}} \right| \leq 1 0 U \sqrt {\bar {t} (\delta / 8)}
$$
Combining the inequality above with (57) we have
$$
\sup  _ {A \in \mathcal {A} _ {p, d}} \sqrt {\mathbb {P} (X \in A)} \left| \mathbb {E} \left(f ^ {*} (X) | X \in A\right) - \bar {y} _ {\mathcal {I} _ {A}} \right| \leq 2 0 U \sqrt {\bar {t} (\delta / 8)} \leq 2 0 U \sqrt {\bar {t} (\delta / 1 2)}
$$
Define  $a \coloneqq \bar{t}(\delta/4)$  and  $b \coloneqq a + \frac{2\delta d}{n}$ . Define events  $\mathcal{E}_1$  and  $\mathcal{E}_2$ :
$$
\mathcal {E} _ {1} := \left\{\max  _ {A \in \tilde {\mathcal {A}} _ {p, d} (b)} \frac {1}{n} \sum_ {i = 1} ^ {n} 1 _ {\{x _ {i} \in A \}} \leq \left(e ^ {2} b\right) \vee \frac {\log \left(2 (n + 1) ^ {2 d} p ^ {d} / \delta\right)}{n} \right\}
$$
$$
\mathcal {E} _ {2} := \left\{\sup  _ {A \in \mathcal {A} _ {p, d} \backslash \mathcal {A} _ {p, d} (a)} \frac {1}{\sqrt {\mathbb {P} (X \in A)}} \Big | \mathbb {P} (X \in A) - \frac {1}{n} \sum_ {i = 1} ^ {n} 1 _ {\{x _ {i} \in A \}} \Big | \leq 5 \sqrt {\bar {t} (\delta / 4)} \right\}
$$
Then by Lemmas A.7 and A.9, we know that  $\mathbb{P}(\mathcal{E}_1) \geq 1 - \delta/2$  and  $\mathbb{P}(\mathcal{E}_2) \geq 1 - \delta/2$ , so  $\mathbb{P}(\mathcal{E}_1 \cap \mathcal{E}_2) \geq 1 - \delta$ . Below we prove (24) when  $\mathcal{E}_1 \cap \mathcal{E}_2$  holds.
For  $A\in \mathcal{A}_{p,d}$ , if  $\mathbb{P}(X\in A)\leq a$ , then  $\mathbb{P}(A^{\prime})\leq a + \frac{2\hat{\theta}d}{n} = b$ . So we have
$$
\begin{array}{l} \sup  _ {A \in \mathcal {A} _ {p, d} (a)} \frac {1}{n} \sum_ {i = 1} ^ {n} 1 _ {\{x _ {i} \in A \}} \leq \sup  _ {A \in \mathcal {A} _ {p, d} (a)} \frac {1}{n} \sum_ {i = 1} ^ {n} 1 _ {\{x _ {i} \in A ^ {\prime} \}} \leq \sup  _ {\tilde {A} \in \tilde {\mathcal {A}} _ {p, d} (b)} \frac {1}{n} \sum_ {i = 1} ^ {n} 1 _ {\{x _ {i} \in \tilde {A} \}} \\ \leq \left(e ^ {2} \bar {t} (\delta / 4) + 2 e ^ {2} \bar {\theta} d / n\right) \vee \frac {\log (2 (n + 1) ^ {2 d} p ^ {d} / \delta)}{n} \tag {62} \\ \leq (e ^ {2} + 1) \bar {t} (\delta / 4) \leq 2 5 \bar {t} (\delta / 4) \\ \end{array}
$$
where the third inequality is by event  $\mathcal{E}_1$  and the definition of  $b$ ; the fourth inequality is because
$$
\bar {t} (\delta / 4) \geq \bar {t} _ {1} (\delta / 4) \geq \frac {1}{n} \log (2 p ^ {d} (n + 1) ^ {2 d} / \delta) \quad \text {a n d} \quad \bar {t} (\delta / 4) \geq \bar {t} _ {2} (\delta / 4) \geq 2 e ^ {2} \bar {\theta} d / n.
$$
$$
\begin{array}{l} \sup  _ {A \in \mathcal {A} _ {p, d} (a)} \left| \sqrt {\mathbb {P} (X \in A)} - \sqrt {\frac {1}{n} \sum_ {i = 1} ^ {n} 1 _ {\{x _ {i} \in A \}}} \right| \tag {63} \\ \leq \sup  _ {A \in \mathcal {A} _ {p, d} (a)} \max  \left\{\sqrt {\mathbb {P} (X \in A)}, \sqrt {\frac {1}{n} \sum_ {i = 1} ^ {n} 1 _ {\{x _ {i} \in A \}}} \right\} \leq 5 \sqrt {\bar {t} (\delta / 4)} \\ \end{array}
$$
where the second inequality made use of (62).
$$
\begin{array}{l} \sup  _ {A \in \mathcal {A} _ {p, d} \backslash \mathcal {A} _ {p, d} (a)} \left| \sqrt {\mathbb {P} (X \in A)} - \sqrt {\frac {1}{n} \sum_ {i = 1} ^ {n} 1 _ {\{x _ {i} \in A \}}} \right| \\ = \sup  _ {A \in \mathcal {A} _ {p, d} \backslash \mathcal {A} _ {p, d} (a)} \frac {\left| \mathbb {P} (X \in A) - \frac {1}{n} \sum_ {i = 1} ^ {n} 1 _ {\{x _ {i} \in A \}} \right|}{\sqrt {\mathbb {P} (X \in A)} + \sqrt {\frac {1}{n} \sum_ {i = 1} ^ {n} 1 _ {\{x _ {i} \in A \}}}} \tag {64} \\ \leq \sup  _ {A \in \mathcal {A} _ {p, d} \backslash \mathcal {A} _ {p, d} (a)} \frac {1}{\sqrt {\mathbb {P} (X \in A)}} \left| \mathbb {P} (X \in A) - \frac {1}{n} \sum_ {i = 1} ^ {n} 1 _ {\{x _ {i} \in A \}} \right| \leq 5 \sqrt {\bar {t} (\delta / 4)} \\ \end{array}
$$
where the last inequality is by event  $\mathcal{E}_2$ .
Combining (63) and (64) the proof is complete.
For any interval  $E \in \mathcal{E}$  and any univariate function  $g$  on  $[0,1]$ , let  $V_{g}(E)$  be the total variation of  $g$  on  $E$ . For the additive model (14) and a rectangle  $A = \prod_{j=1}^{p} E_{j} \in \mathcal{A}$ , define  $V_{f^{*}}(A) = \sum_{j=1}^{p} V_{f_{j}^{*}}(E_{j})$ . Recall that  $X$  is a random variable with the same distribution as  $x_{i}$ , and  $X^{(j)}$  is the  $j$ -th coordinate of  $X$ .
Lemma B.1 For any rectangle  $A \subseteq [0,1]^p$ , any  $j \in [p]$  and any  $b \in \mathbb{R}$ , it holds
$$
\Delta (A, j, b) = \left(\mathbb {E} (f ^ {*} (X) 1 _ {\{X \in A _ {R} \}}) - \mathbb {E} (f ^ {*} (X) | X \in A) \mathbb {P} (X \in A _ {R})\right) ^ {2} \frac {\mathbb {P} (X \in A)}{\mathbb {P} (X \in A _ {L}) \mathbb {P} (X \in A _ {R})}
$$
where  $A_{L} = A_{L}(j,b)$  and  $A_{R} = A_{R}(j,b)$ .
Proof. We use the notations  $\nu \coloneqq \mathbb{E}(f^{*}(X)|X\in A)$ ,  $\nu_{L}\coloneqq \mathbb{E}(f^{*}(X)|X\in A_{L})$  and  $\nu_{R}\coloneqq \mathbb{E}(f^{*}(X)|X\in A_{R})$ . First, note that
$$
\begin{array}{l} \mathbb {E} \left(\left(f ^ {*} (X) - \nu\right) ^ {2} 1 _ {\{X \in A _ {L} \}}\right) = \mathbb {E} \left(\left(f ^ {*} (X) - \nu_ {L} + \nu_ {L} - \nu\right) ^ {2} 1 _ {\{X \in A _ {L} \}}\right) \\ = \mathbb {E} \left(\left(f ^ {*} (X) - \nu_ {L}\right) ^ {2} 1 _ {\{X \in A _ {L} \}}\right) + (\nu_ {L} - \nu) ^ {2} \mathbb {P} (X \in A _ {L}) \tag {05} \\ \end{array}
$$
$$
\mathbb {E} \left(\left(f ^ {*} (X) - \nu\right) ^ {2} 1 _ {\{X \in A _ {R} \}}\right) = \mathbb {E} \left(\left(f ^ {*} (X) - \nu_ {R}\right) ^ {2} 1 _ {\{X \in A _ {R} \}}\right) + (\nu_ {R} - \nu) ^ {2} \mathbb {P} (X \in A _ {R}) \tag {66}
$$
Summing up (65) and (66) we have
$$
\begin{array}{l} \left(\nu_ {L} - \nu\right) ^ {2} \mathbb {P} \left(X \in A _ {L}\right) + \left(\nu_ {R} - \nu\right) ^ {2} \mathbb {P} \left(X \in A _ {R}\right) \\ = \mathbb {E} \left(\left(f ^ {*} (X) - \nu\right) ^ {2} 1 _ {\{X \in A \}}\right) - \mathbb {E} \left(\left(f ^ {*} (X) - \nu_ {L}\right) ^ {2} 1 _ {\{X \in A _ {L} \}}\right) - \mathbb {E} \left(\left(f ^ {*} (X) - \nu_ {R}\right) ^ {2} 1 _ {\{X \in A _ {R} \}}\right) \tag {67} \\ = \Delta (A, j, b) \\ \end{array}
$$
$$
\begin{array}{l} \left(\nu_ {L} - \nu\right) ^ {2} \mathbb {P} (X \in A _ {L}) = \left(\mathbb {E} \left(f ^ {*} (X) 1 _ {\{X \in A _ {L} \}}\right) - \nu \mathbb {P} (X \in A _ {L})\right) ^ {2} \left(\mathbb {P} (X \in A _ {L})\right) ^ {- 1} \\ = \left(\nu \mathbb {P} (X \in A) - \mathbb {E} \left(f ^ {*} (X) 1 _ {\{X \in A _ {R} \}}\right) - \nu \mathbb {P} (X \in A _ {L})\right) ^ {2} (\mathbb {P} (X \in A _ {L})) ^ {- 1} \\ = \left(\nu \mathbb {P} (X \in A _ {R}) - \mathbb {E} \left(f ^ {*} (X) 1 _ {\{X \in A _ {R} \}}\right)\right) ^ {2} (\mathbb {P} (X \in A _ {L})) ^ {- 1} \tag {68} \\ = (\nu_ {R} - \nu) ^ {2} \frac {\left(\mathbb {P} (X \in A _ {R})\right) ^ {2}}{\mathbb {P} (X \in A _ {L})} \\ \end{array}
$$
Combining (67) and (68) we have
$$
\begin{array}{l} \Delta (A, j, b) = (\nu_ {R} - \nu) ^ {2} \frac {\left(\mathbb {P} (X \in A _ {R})\right) ^ {2}}{\mathbb {P} (X \in A _ {L})} + (\nu_ {R} - \nu) ^ {2} \mathbb {P} (X \in A _ {R}) = (\nu_ {R} - \nu) ^ {2} \frac {\mathbb {P} (X \in A _ {R}) \mathbb {P} (X \in A)}{\mathbb {P} (X \in A _ {L})} \\ = \left(\mathbb {E} (f ^ {*} (X) 1 _ {\{X \in A _ {R} \}}) - \nu \mathbb {P} (X \in A _ {R})\right) ^ {2} \frac {\mathbb {P} (X \in A)}{\mathbb {P} (X \in A _ {L}) \mathbb {P} (X \in A _ {R})} \\ \end{array}
$$
Lemma B.2 Suppose Assumption 2.1 holds true, and  $f^*$  has the additive structure in (14). Then for any  $A = \prod_{j=1}^{p} [\ell_j, u_j] \subseteq [0,1]^p$ , it holds
$$
\max  _ {j \in [ p ], b \in \mathbb {R}} \sqrt {\Delta (A , j , b)} \geq \frac {\sqrt {\mathbb {P} (X \in A)} \operatorname {V a r} (f ^ {*} (X) | X \in A)}{\sum_ {k = 1} ^ {p} \int_ {\ell_ {k}} ^ {u _ {k}} \sqrt {q _ {A} ^ {(k)} (t) (1 - q _ {A} ^ {(k)} (t))} d V _ {f _ {k} ^ {*}} ([ \ell_ {j} , t ])}
$$
where  $q_A^{(k)}(t)\coloneqq \mathbb{P}(X^{(k)}\leq t|x_1\in A)$
Proof. For a fixed  $A = \prod_{j=1}^{p} [\ell_j, u_j] \subseteq [0,1]^p$ , without loss of generality, assume  $\mathbb{E}(f^*(X) | X \in A) = 0$ . Note that for any  $j \in [p]$ ,
$$
\max  _ {b \in \mathbb {R}} \sqrt {\Delta (A , j , b)} \geq \frac {\int_ {\ell_ {j}} ^ {u _ {j}} \sqrt {q _ {A} ^ {(j)} (s) \left(1 - q _ {A} ^ {(j)} (s)\right)} \sqrt {\Delta (A , j , s)} d V _ {f _ {j} ^ {*}} \left(\left[ \ell_ {j} , s \right]\right)}{\int_ {\ell_ {j}} ^ {u _ {j}} \sqrt {q _ {A} ^ {(j)} (s) \left(1 - q _ {A} ^ {(j)} (s)\right)} d V _ {f _ {j} ^ {*}} \left(\left[ \ell_ {j} , s \right]\right)} \tag {69}
$$
where  $s$  the integration variable. Because  $q_A^{(j)}(s) = \mathbb{P}(X\in A_L(j,s)) / \mathbb{P}(X\in A)$ , using Lemma B.1 and recall that we have assumed  $\mathbb{E}(f^{*}(X)|X\in A) = 0$ , we have
$$
\begin{array}{l} \int_ {\ell_ {j}} ^ {u _ {j}} \sqrt {q _ {A} ^ {(j)} (s) (1 - q _ {A} ^ {(j)} (s))} \sqrt {\Delta (A , j , s)} d V _ {f _ {j} ^ {*}} ([ \ell_ {j}, s ]) \\ = \frac {1}{\sqrt {\mathbb {P} (X \in A)}} \int_ {\ell_ {j}} ^ {u _ {j}} \left| \mathbb {E} (f ^ {*} (X) 1 _ {\{X \in A _ {R} (j, s) \}}) \right| d V _ {f _ {j} ^ {*}} ([ \ell_ {j}, s ]) \\ = \frac {1}{\sqrt {\mathbb {P} (X \in A)}} \int_ {\ell_ {j}} ^ {u _ {j}} \left| \mathbb {E} \left(f ^ {*} (X) 1 _ {\{X \in A \}} 1 _ {\{X ^ {(j)} > s \}}\right) \right| d V _ {f _ {j} ^ {*}} ([ \ell_ {j}, s ]) \\ \geq \frac {1}{\sqrt {\mathbb {P} (X \in A)}} \left| \int_ {\ell_ {j}} ^ {u _ {j}} \mathbb {E} \left(f ^ {*} (X) 1 _ {\{X \in A \}} 1 _ {\{X ^ {(j)} > s \}}\right) d f _ {j} ^ {*} (s) \right| \\ = \frac {1}{\sqrt {\mathbb {P} (X \in A)}} \left| \mathbb {E} \left(f ^ {*} (X) \left(f _ {j} ^ {*} \left(X ^ {(j)}\right) - f _ {j} ^ {*} (\ell_ {j})\right) 1 _ {\{X \in A \}}\right) \right| \\ = \frac {1}{\sqrt {\mathbb {P} (X \in A)}} \left| \mathbb {E} \left(f ^ {*} (X) f _ {j} ^ {*} (X ^ {(j)}) 1 _ {\{X \in A \}}\right) \right| \\ \end{array}
$$
where the last equality makes use of the assumption that  $\mathbb{E}(f^{*}(X)|X\in A) = 0$ . Combining the inequality above with (69), we have
$$
\max  _ {b \in \mathbb {R}} \sqrt {\Delta (A , j , b)} \geq \frac {1}{\sqrt {\mathbb {P} (X \in A)}} \frac {\left| \mathbb {E} (f ^ {*} (X) f _ {j} ^ {*} (X ^ {(j)}) 1 _ {\{X \in A \}})\right|}{\int_ {\ell_ {j}} ^ {u _ {j}} \sqrt {q _ {A} ^ {(j)} (s) (1 - q _ {A} ^ {(j)} (s))} d V _ {f _ {j} ^ {*}} ([ \ell_ {j} , s ])}
$$
$$
\max  _ {j \in [ p ], b \in \mathbb {R}} \sqrt {\Delta (A , j , b)} \geq \frac {1}{\sqrt {\mathbb {P} (X \in A)}} \frac {\sum_ {j = 1} ^ {p} \left| \mathbb {E} \left(f ^ {*} (X) f _ {j} ^ {*} \left(X ^ {(j)}\right) 1 _ {\{X \in A \}}\right) \right|}{\sum_ {j = 1} ^ {p} \int_ {\ell_ {j}} ^ {u _ {j}} \sqrt {q _ {A} ^ {(j)} (s) \left(1 - q _ {A} ^ {(j)} (s)\right)} d V _ {f _ {j} ^ {*}} ([ \ell_ {j} , s ])} \tag {70}
$$
By the additive structure (14) we have
$$
\sum_ {j = 1} ^ {p} \left| \mathbb {E} \left(f ^ {*} (X) f _ {j} ^ {*} \left(X ^ {(j)}\right) 1 _ {\{X \in A \}}\right) \right| \geq \mathbb {E} \left(\left(f ^ {*} (X)\right) ^ {2} 1 _ {\{X \in A \}}\right) = \mathbb {P} (X \in A) \operatorname {V a r} \left(f ^ {*} (X) | X \in A\right) \tag {71}
$$
Combining (70) and (71), the proof is complete.
Lemma B.3 Suppose Assumption 2.1 holds true, and  $f^{*}$  has the additive structure in (14). If for any  $A = \prod_{j=1}^{p} [\ell_j, u_j] \subseteq [0,1]^p$  and any  $k \in [p]$  it holds
$$
\left(\int_ {\ell_ {k}} ^ {u _ {k}} \sqrt {q _ {A} ^ {(k)} (t) \left(1 - q _ {A} ^ {(k)} (t)\right)} d V _ {f _ {k} ^ {*}} ([ \ell_ {k}, t ])\right) ^ {2} \leq \frac {\tau^ {2}}{u _ {k} - \ell_ {k}} \inf  _ {w \in \mathbb {R}} \int_ {\ell_ {k}} ^ {u _ {k}} | f _ {k} ^ {*} (t) - w | ^ {2} d t \tag {72}
$$
Then Assumption 2.2 is satisfied with  $\lambda = \theta / (p\tau^2\bar{\theta})$ .
Proof. Given  $A = \prod_{j=1}^{p} [\ell_j, u_j] \subseteq [0,1]^p$ , without loss of generality, assume  $\mathbb{E}(f^*(X) | X \in A) = 0$ . Let  $p_X(\cdot)$  be the density of  $X$  on  $[0,1]^p$ . Then we have
$$
\operatorname {V a r} \left(f ^ {*} (X) \mid X \in A\right) = \frac {1}{\mathbb {P} (X \in A)} \int_ {A} \left(f ^ {*} (z)\right) ^ {2} p _ {X} (z) d z \geq \frac {\theta}{\mathbb {P} (X \in A)} \int_ {A} \left(f ^ {*} (z)\right) ^ {2} d z \tag {73}
$$
where the second inequality made use of Assumption 2.1 (i). Denote  $c_{j} \coloneqq \frac{1}{u_{j} - \ell_{j}}\int_{\ell_{j}}^{u_{j}}f_{j}^{*}(t)dt$  and  $c \coloneqq \sum_{j = 1}^{p}c_{j}$ , then we have
$$
\begin{array}{l} \int_ {A} \left(f ^ {*} (z)\right) ^ {2} d z = \int_ {A} \left(c + \sum_ {j = 1} ^ {p} f _ {j} ^ {*} \left(z _ {j}\right) - c _ {j}\right) ^ {2} d z _ {1} d z _ {2} \dots d z _ {p} \\ = \int_ {A} c ^ {2} + \sum_ {j = 1} ^ {p} \left(f _ {j} ^ {*} \left(z _ {j}\right) - c _ {j}\right) ^ {2} d z _ {1} d z _ {2} \dots d z _ {p} \\ \geq \sum_ {j = 1} ^ {p} \frac {\prod_ {k = 1} ^ {p} \left(u _ {k} - \ell_ {k}\right)}{u _ {j} - \ell_ {j}} \int_ {\ell_ {j}} ^ {u _ {j}} \left(f _ {j} ^ {*} (t) - c _ {j}\right) ^ {2} d t \\ \geq \frac {\mathbb {P} (X \in A)}{\bar {\theta}} \sum_ {j = 1} ^ {p} \frac {1}{u _ {j} - \ell_ {j}} \int_ {\ell_ {j}} ^ {u _ {j}} \left(f _ {j} ^ {*} (t) - c _ {j}\right) ^ {2} d t \\ \end{array}
$$
Combining the inequality above with (73) we have
$$
\operatorname {V a r} \left(f ^ {*} (X) \mid X \in A\right) \geq \frac {\theta}{\bar {\theta}} \sum_ {j = 1} ^ {p} \frac {1}{u _ {j} - \ell_ {j}} \int_ {\ell_ {j}} ^ {u _ {j}} \left(f _ {j} ^ {*} (t) - c _ {j}\right) ^ {2} d t \tag {74}
$$
We use  $H_{k}^{2}$  to denote the LHS of (72), then (72) implies
$$
\frac {1}{u _ {j} - \ell_ {j}} \int_ {\ell_ {j}} ^ {u _ {j}} \left| f _ {j} ^ {*} (t) - c _ {j} \right| ^ {2} d t \geq \frac {1}{\tau^ {2}} H _ {j} ^ {2} \tag {75}
$$
As a result of (74) and (75), we have
$$
\operatorname {V a r} \left(f ^ {*} (X) \mid X \in A\right) \geq \frac {\underline {{\theta}}}{\bar {\theta} \tau^ {2}} \sum_ {j = 1} ^ {p} H _ {k} ^ {2} \tag {76}
$$
By Lemma B.2 we have
$$
\begin{array}{l} \max  _ {j \in [ p ], b \in \mathbb {R}} \Delta (A, j, b) \geq \frac {\mathbb {P} (X \in A) \operatorname {V a r} \left(f ^ {*} (X) \mid X \in A\right) ^ {2}}{\left(\sum_ {k = 1} ^ {p} H _ {k}\right) ^ {2}} \\ \geq \frac {\theta}{\bar {\theta} \tau^ {2}} \frac {\sum_ {j = 1} ^ {p} H _ {k} ^ {2}}{\left(\sum_ {k = 1} ^ {p} H _ {k}\right) ^ {2}} \mathbb {P} (X \in A) \operatorname {V a r} \left(f ^ {*} (X) | X \in A\right) \\ \geq \frac {\theta}{p \bar {\theta} \tau^ {2}} \mathbb {P} (X \in A) \operatorname {V a r} \left(f ^ {*} (X) | X \in A\right) \\ \end{array}
$$
where the second inequality is by (76), and the last inequality made use of the Cauchy-Schwarz inequality.
For any  $A = \prod_{j=1}^{p}[\ell_j, u_j] \subseteq [0,1]^p$  and any  $k \in [p]$  it holds
$$
\begin{array}{l} \left(\int_ {\ell_ {k}} ^ {u _ {k}} \sqrt {q _ {A} ^ {(k)} (t) \left(1 - q _ {A} ^ {(k)} (t)\right)} d V _ {f _ {k} ^ {*}} ([ \ell_ {k}, t ])\right) ^ {2} \leq \frac {1}{4} \left(\int_ {\ell_ {k}} ^ {u _ {k}} | \left(f _ {k} ^ {*}\right) ^ {\prime} (t) | d t\right) ^ {2} \\ \leq \frac {\tau^ {2} / 4}{u _ {k} - \ell_ {k}} \inf  _ {w \in \mathbb {R}} \int_ {\ell_ {k}} ^ {u _ {k}} | f _ {k} ^ {*} (t) - w | ^ {2} \mathrm {d} t \\ \end{array}
$$
where the first inequality is by Cauchy-Schwarz inequality, and the second is because  $f_{k}^{*}\in LRP([0,1],\tau)$ .
Using Lemma B.3, the proof of complete.
For any  $A = \prod_{j=1}^{p}[\ell_j, u_j] \subseteq [0,1]^p$  and any  $k \in [p]$ , we prove that
$$
\left(\int_ {\ell_ {k}} ^ {u _ {k}} \sqrt {q _ {A} ^ {(k)} (t) \left(1 - q _ {A} ^ {(k)} (t)\right)} d V _ {f _ {k} ^ {*}} ([ \ell_ {k}, t ])\right) ^ {2} \leq \max  \left\{\frac {2 r \bar {\theta}}{\theta}, \frac {r ^ {2}}{2 \alpha} \right\} \frac {\max  \left\{9 \beta^ {2} , 3 2 + \beta^ {2} \right\}}{u _ {k} - \ell_ {k}} \inf  _ {w \in \mathbb {R}} \int_ {\ell_ {k}} ^ {u _ {k}} | f _ {k} ^ {*} (t) - w | ^ {2} d t \tag {77}
$$
Then the conclusion follows Lemma B.3.
For fixed  $A$  and  $k \in [p]$ , to simplify the notation, we denote  $g := f_k^*$ ,  $a := \ell_k$ ,  $b := u_k$ ,  $q(t) := q_A^{(k)}(t)$  for all  $t \in [\ell_k, u_k]$ , and  $t_j := t_j^{(k)}$  for  $j = 0, 1, \dots, r$ . Then (77) can be written as
$$
\left(\int_ {a} ^ {b} \sqrt {q (t) (1 - q (t))} d V _ {g} ([ a, t ])\right) ^ {2} \leq 2 r \max  \left\{\frac {\bar {\theta}}{\underline {{\theta}}}, \frac {r}{4 \alpha} \right\} \frac {\max  \left\{9 \beta^ {2} , 3 2 + \beta^ {2} \right\}}{b - a} \inf  _ {w \in \mathbb {R}} \int_ {a} ^ {b} (g (t) - w) ^ {2} d t \tag {78}
$$
For any  $s \in (0,1)$ , define  $\Delta g(s) \coloneqq \lim_{t \to s+} g(t) - \lim_{t \to s-} g(t)$ . Let  $j', j'' \in [r]$  such that  $t_{j' - 1} \leq a < t_{j'}$  and  $t_{j'' - 1} < b \leq t_{j''}$ , and define  $r' = j'' - j' + 1$ , and
$$
z _ {0} = a, z _ {1} = t _ {j ^ {\prime}}, z _ {2} = t _ {j ^ {\prime} + 1}, \dots , z _ {r ^ {\prime} - 1} = t _ {j ^ {\prime \prime} - 1}, z _ {r ^ {\prime}} = b.
$$
$$
\begin{array}{l} \left(\int_ {a} ^ {b} \sqrt {q (t) (1 - q (t))} d V _ {g} ([ a, t ])\right) ^ {2} \\ = \left(\sum_ {j = 1} ^ {r ^ {\prime}} \int_ {z _ {j - 1}} ^ {z _ {j}} \sqrt {q (t) (1 - q (t))} | g ^ {\prime} (t) | \mathrm {d} t + \sum_ {j = 1} ^ {r ^ {\prime} - 1} \sqrt {q \left(z _ {j}\right) \left(1 - q \left(z _ {j}\right)\right)} \Delta g \left(z _ {j}\right)\right) ^ {2} \tag {79} \\ \leq 2 r ^ {\prime} \sum_ {j = 1} ^ {r ^ {\prime}} \left(\int_ {z _ {j - 1}} ^ {z _ {j}} \sqrt {q (t) (1 - q (t))} | g ^ {\prime} (t) | d t\right) ^ {2} + 2 \left(r ^ {\prime} - 1\right) \sum_ {j = 1} ^ {r ^ {\prime} - 1} q \left(z _ {j}\right) \left(1 - q \left(z _ {j}\right)\right) \left| \Delta g \left(z _ {j}\right) \right| ^ {2} \\ \end{array}
$$
We have the following 4 claims bounding the terms in the last line of the display above.
Claim B.4 For  $j \in \{1, r'\}$ , it holds
$$
\left(\int_ {z _ {j - 1}} ^ {z _ {j}} \sqrt {q (t) (1 - q (t))} | g ^ {\prime} (t) | \mathrm {d} t\right) ^ {2} \leq \frac {\bar {\theta} \beta^ {2}}{\underline {{\theta}} (b - a)} \inf  _ {w \in \mathbb {R}} \int_ {z _ {j - 1}} ^ {z _ {j}} (g (t) - w) ^ {2} \mathrm {d} t
$$
Proof of Claim B.4: We just prove the claim for  $j = 1$ . The proof for  $j = r'$  follows a similar argument. To prove the claim for  $j = 1$ , we discuss two cases.
$\underline{\mathrm{(Case~1)}} q(z_1)\leq 1 / 2$  . Then we have  $\sqrt{q(t)(1 - q(t))}\leq \sqrt{q(z_1)(1 - q(z_1))}$  , hence
$$
\begin{array}{l} \left(\int_ {a} ^ {z _ {1}} \sqrt {q (t) (1 - q (t))} | g ^ {\prime} (t) | \mathrm {d} t\right) ^ {2} \leq q \left(z _ {1}\right) \left(1 - q \left(z _ {1}\right)\right) \left(\int_ {a} ^ {z _ {1}} | g ^ {\prime} (t) | \mathrm {d} t\right) ^ {2} \\ \leq q \left(z _ {1}\right) \left(1 - q \left(z _ {1}\right)\right) \frac {\beta^ {2}}{z _ {1} - a} \inf  _ {w \in \mathbb {R}} \int_ {a} ^ {z _ {1}} (g (t) - w) ^ {2} d t \tag {80} \\ \leq \frac {\bar {\theta} \beta^ {2}}{\bar {\theta} (b - a)} \inf  _ {w \in \mathbb {R}} \int_ {a} ^ {z _ {1}} (g (t) - w) ^ {2} d t \\ \end{array}
$$
where the second inequality is because  $g \in LRP((a, z_1), \beta)$ ; and the last inequality makes use of the fact  $q(z_1) \leq \bar{\theta}(z_1 - a) / (\underline{\theta}(b - a))$ .
Case 2)  $q(z_{1}) > 1 / 2$ . Then we have
$$
z _ {1} - a \geq \frac {\theta (b - a)}{\bar {\theta}} q \left(z _ {1}\right) \geq \frac {\theta (b - a)}{2 \bar {\theta}} \tag {81}
$$