SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \int_ {B} g (\omega) \mathbb {P} ^ {\mathrm {d o} (U, \mathbb {Q})} (d \omega) = \int_ {B} \lim  _ {n \rightarrow \infty} f _ {n} (\omega) \mathbb {P} ^ {\mathrm {d o} (U, \mathbb {Q})} (d \omega) \\ = \lim  _ {n \rightarrow \infty} \int_ {B} f _ {n} (\omega) \mathbb {P} ^ {\mathrm {d o} (U, \mathbb {Q})} (d \omega) \quad \text {b y} \\ \end{array}
$$
$$
\begin{array}{l} = \lim  _ {n \rightarrow \infty} \int_ {B} f _ {n} (\omega) \mathbb {Q} (d \omega) \quad \text {b y} \\ = \int_ {B} \lim  _ {n \rightarrow \infty} f _ {n} (\omega) \mathbb {Q} (d \omega) \quad \text {b y} \\ = \int_ {B} g (\omega) \mathbb {Q} (d \omega). \\ \end{array}
$$
We use this fact in the proof of both parts of this theorem.
(i) First note that we indeed have  $K_U^{\mathrm{do}(U,\mathbb{Q},\mathbb{L})} = K_U$ , by Remark C.1(a). For any  $A \in \mathcal{H}$ , the map  $\omega \mapsto K_U(\omega, A)$  is  $\mathcal{H}_U$ -measurable, so for any  $B \in \mathcal{H}_U$ ,
$$
\begin{array}{l} \int_ {B} K _ {U} (\omega , A) \mathbb {P} ^ {\mathrm {d o} (U, \mathbb {Q})} (d \omega) = \int_ {B} K _ {U} (\omega , A) \mathbb {Q} (d \omega) \quad \text {b y} \\ = \int 1 _ {B} (\omega) K _ {U} (\omega , A) \mathbb {Q} (d \omega) \\ = \int K _ {U} (\omega , A \cap B) \mathbb {Q} (d \omega) \quad \text {b y} \\ = \mathbb {P} ^ {\mathrm {d o} (U, \mathbb {Q})} (A \cap B) \\ = \int 1 _ {A \cap B} (\omega) \mathbb {P} ^ {\mathrm {d o} (U, \mathbb {Q})} (d \omega) \\ = \int 1 _ {B} (\omega) 1 _ {A} (\omega) \mathbb {P} ^ {\mathrm {d o} (U, \mathbb {Q})} (d \omega) \\ = \int_ {B} 1 _ {A} (\omega) \mathbb {P} ^ {\mathrm {d o} (U, \mathbb {Q})} (d \omega). \\ \end{array}
$$
So  $K_U(\cdot, A) = K_U^{\mathrm{do}(U, \mathbb{Q}, \mathbb{L})}(\cdot, A)$  is indeed a version of the conditional probability  $\mathbb{P}_{\mathcal{H}_U}^{\mathrm{do}(U, \mathbb{Q})}(A)$ , which means that  $\mathcal{H}_U$  is a global source of  $(\Omega, \mathcal{H}, \mathbb{P}^{\mathrm{do}(U, \mathbb{Q})}, \mathbb{K}^{\mathrm{do}(U, \mathbb{Q}, \mathbb{L})})$ .
(ii) For any  $A \in \mathcal{H}$ , the map  $\omega \mapsto K_V^{\mathrm{do}(U,\mathbb{Q})}(\omega,A)$  is  $\mathcal{H}_V$ -measurable and hence  $\mathcal{H}_U$ -measurable, so for any  $B \in \mathcal{H}_V \subseteq \mathcal{H}_U$ ,
$$
\begin{array}{l} \int_ {B} K _ {V} ^ {\mathrm {d o} (U, \mathbb {Q})} (\omega_ {V}, A) \mathbb {P} ^ {\mathrm {d o} (U, \mathbb {Q})} (d \omega_ {V}) \\ = \int_ {B} K _ {V} ^ {\mathrm {d o} (U, \mathbb {Q})} \left(\omega_ {V}, A\right) \mathbb {Q} \left(d \omega_ {V}\right) \quad \text {b y} \\ = \int K _ {V} ^ {\mathrm {d o} (U, \mathbb {Q})} \left(\omega_ {V}, A \cap B\right) \mathbb {Q} \left(d \omega_ {V}\right) \quad \text {b y} \\ = \int \int \mathbb {Q} \left(d \omega_ {U \backslash V} ^ {\prime}\right) K _ {U} \left(\left(\omega_ {V}, \omega_ {U \backslash V} ^ {\prime}\right), A \cap B\right) \mathbb {Q} \left(d \omega_ {V}\right) \\ = \int K _ {U} \left(\omega_ {U}, A \cap B\right) \mathbb {Q} \left(d \omega_ {U}\right) \\ = \int_ {B} 1 _ {A} (\omega) \mathbb {P} ^ {\mathrm {d o} (U, \mathbb {Q})} (d \omega). \\ \end{array}
$$
where, in going from the third line to the fourth, we used Theorem C.3, and to go from the fourth line to the fifth, we used the hypothesis that  $\mathbb{Q}$  factorises over  $\mathcal{H}_V$  and  $\mathcal{H}_{U\backslash V}$ , meaning  $\mathbb{Q}(d\omega_{U\backslash V})\mathbb{Q}(d\omega_V) = \mathbb{Q}(d\omega_U)$ . So  $K_V^{\mathrm{do}(U,\mathbb{Q})}(\omega ,A)$  is indeed a version of the conditional probability  $\mathbb{P}_{\mathcal{H}_V}^{\mathrm{do}(U,\mathbb{Q})}(A)$ , which means that  $\mathcal{H}_V$  is a global source of  $(\Omega ,\mathcal{H},\mathbb{P}^{\mathrm{do}(U,\mathbb{Q})},\mathbb{K}^{\mathrm{do}(U,\mathbb{Q})})$ .
(i) For  $A \in \mathcal{H}_U$  and  $V \in \mathcal{P}(T)$  with  $V \cap U = \emptyset$ ,  $\mathcal{H}_V$  has no causal effect on  $A$  (c.f. Definition B.1(i)) in the intervention causal space  $(\Omega, \mathcal{H}, \mathbb{P}^{\mathrm{do}(U,\mathbb{Q})}, \mathbb{K}^{\mathrm{do}(U,\mathbb{Q},\mathbb{L})})$ , i.e. events in the  $\sigma$ -algebra  $\mathcal{H}_U$  on which intervention took place are not causally affected by  $\sigma$ -algebras outside  $\mathcal{H}_U$ .
(ii) Again, let  $V \in \mathcal{P}(T)$  with  $V \cap U = \emptyset$ , and also let  $A \in \mathcal{H}$  be any event. If, in the original causal space,  $\mathcal{H}_V$  had no causal effect on  $A$ , then in the intervention causal space,  $\mathcal{H}_V$  has no causal effect on  $A$  either.
(iii) Now let  $V \in \mathcal{P}(T)$ ,  $A \in \mathcal{H}$  any event and suppose that the intervention on  $\mathcal{H}_U$  via  $\mathbb{Q}$  is hard. Then if  $\mathcal{H}_V$  had no causal effect on  $A$  in the original causal space, then  $\mathcal{H}_V$  has no causal effect on  $A$  in the intervention causal space either.
Proof. (i) Take any  $S \in \mathcal{P}(T)$ . See that
$$
\begin{array}{l} K _ {S} ^ {\mathrm {d o} (U, \mathbb {Q}, \mathbb {L})} (\omega , A) = \int L _ {S \cap U} \left(\omega_ {S \cap U}, d \omega_ {U} ^ {\prime}\right) K _ {S \cup U} \left(\left(\omega_ {S \setminus U}, \omega_ {U} ^ {\prime}\right), A\right) \\ = \int L _ {S \cap U} (\omega_ {S \cap U}, d \omega_ {U} ^ {\prime}) 1 _ {A} (\omega_ {U} ^ {\prime}) \\ = \int L _ {S \cap U} (\omega_ {S \cap U}, d \omega_ {U} ^ {\prime}) K _ {(S \setminus V) \cup U} ((\omega_ {(S \setminus V) \setminus U}, \omega_ {U} ^ {\prime}), A) \\ = \int L _ {(S \backslash V) \cap U} (\omega_ {(S \backslash V) \cap U}, d \omega_ {U} ^ {\prime}) K _ {(S \backslash V) \cup U} ((\omega_ {(S \backslash V) \backslash U}, \omega_ {U} ^ {\prime}), A) \\ = K _ {S \backslash V} ^ {\mathrm {d o} (U, \mathbb {Q}, \mathbb {L})} (\omega , A) \\ \end{array}
$$
where, in going from the first line to the second and from the second line to the third, we used the fact that  $A \in \mathcal{H}_U$ , and in going from the third line to the fourth, we applied the fact that  $(S \setminus V) \cap U = S \cap U$  since  $V \cap U = \emptyset$ . Since  $S \in \mathcal{P}(T)$  was arbitrary,  $\mathcal{H}_V$  has no causal effect on  $A$  in the intervention causal space.
(ii) Take any  $S\in \mathcal{P}(T)$  . See that
$$
\begin{array}{l} K _ {S} ^ {\mathrm {d o} (U, \mathbb {Q}, \mathbb {L})} (\omega , A) = \int L _ {S \cap U} \left(\omega_ {S \cap U}, d \omega_ {U} ^ {\prime}\right) K _ {S \cup U} \left(\left(\omega_ {S \setminus U}, \omega_ {U} ^ {\prime}\right), A\right) \\ = \int L _ {S \cap U} \left(\omega_ {S \cap U}, d \omega_ {U} ^ {\prime}\right) K _ {(S \cup U) \backslash V} \left(\left(\omega_ {(S \backslash V) \backslash U}, \omega_ {U} ^ {\prime}\right), A\right) \\ = \int L _ {(S \backslash V) \cap U} \left(\omega_ {(S \backslash V) \cap U}, d \omega_ {U} ^ {\prime}\right) K _ {(S \backslash V) \cup U} \left(\left(\omega_ {(S \backslash V) \backslash U}, \omega_ {U} ^ {\prime}\right), A\right) \\ = K _ {S \backslash V} ^ {\mathrm {d o} (U, \mathbb {Q}, \mathbb {L})} (\omega , A) \\ \end{array}
$$
where, in going from the first line to the second, we used the fact that  $\mathcal{H}_V$  has no causal effect on  $A$  in the original causal space, and in going from the second line to the third, we used  $U\cap V = \emptyset$ , which gives us  $S\cap U = (S\setminus V)\cap U$  and  $(S\cup U)\setminus V = (S\setminus V)\cup U$ . Since  $S\in \mathbb{P}(T)$  was arbitrary,  $\mathcal{H}_V$  has no causal effect on  $A$  in the intervention causal space.
(iii) Take any  $S \in \mathcal{P}(T)$ . Apply Theorem C.3 to see that
$$
\begin{array}{l} K _ {S} ^ {\mathrm {d o} (U, \mathbb {Q}, \mathrm {h a r d})} (\omega , A) = \int \mathbb {Q} (d \omega_ {U \setminus S} ^ {\prime}) K _ {S \cup U} ((\omega_ {S}, \omega_ {U \setminus S} ^ {\prime}), A) \\ = \int \mathbb {Q} \left(d \omega_ {U \backslash S} ^ {\prime}\right) K _ {(S \cup U) \backslash V} \left(\left(\omega_ {S}, \omega_ {U \backslash S} ^ {\prime}\right), A\right) \quad \text {D e f .} B. 1 (i) \\ = \int \mathbb {Q} \left(d \omega_ {U \backslash S} ^ {\prime}\right) K _ {\left(\left(S \backslash V\right) \cup U\right) \backslash V} \left(\left(\omega_ {S}, \omega_ {U \backslash S} ^ {\prime}\right), A\right) \\ = \int \mathbb {Q} \left(d \omega_ {U \backslash S} ^ {\prime}\right) K _ {(S \backslash V) \cup U} \left(\left(\omega_ {S}, \omega_ {U \backslash S} ^ {\prime}\right), A\right) \quad \text {D e f .} B. 1 (i) \\ = \int \mathbb {Q} \left(d \omega_ {U \backslash (S \backslash V)} ^ {\prime}\right) K _ {(S \backslash V) \cup U} \left(\left(\omega_ {S \backslash V}, \omega_ {U \backslash (S \backslash V)} ^ {\prime}\right), A\right) \\ \end{array}
$$
$$
= K _ {S \backslash V} ^ {\mathrm {d o} (U, \mathbb {Q})} (\omega , A),
$$
where, in going from the second line to the third, we used that  $(S\cup U)\backslash V = ((S\backslash V)\cup U)\backslash V$  Since  $S\in \mathcal{P}(T)$  was arbitrary,  $\mathcal{H}_V$  has no causal effect on  $A$  in the intervention causal space.
Proof. That  $\mathcal{H}_U$  has a dormant causal effect on  $A$  tells us that  $K_U(\omega, A) = \mathbb{P}(A)$  for all  $\omega \in \Omega$ , but there exists some  $S \in \mathcal{P}(T)$  and some  $\omega_0 \in \Omega$  such that  $K_S(\omega_0, A) \neq K_{S \setminus U}(\omega_0, A)$ . We must have  $S \cap U \neq \emptyset$ , since otherwise  $S \setminus U = S$  and we cannot possibly have  $K_S(\omega_0, A) \neq K_{S \setminus U}(\omega_0, A)$ . Then we hard-intervene on  $\mathcal{H}_{S \setminus U}$  with the Dirac measure on  $\omega_0$ . Then apply Theorem C.3 to see that
$$
\begin{array}{l} K _ {S \cap U} ^ {\mathrm {d o} (S \backslash U, \delta_ {\omega_ {0}}, \text {h a r d})} ((\omega_ {0}) _ {U \cap S}, A) = \int \delta_ {\omega_ {0}} (d \omega_ {S \backslash U} ^ {\prime}) K _ {S} (((\omega_ {0}) _ {U \cap S}, \omega_ {S \backslash U} ^ {\prime}), A) \\ = K _ {S} \left(\omega_ {0}, A\right) \\ \neq K _ {S \backslash U} (\omega_ {0}, A) \\ \end{array}
$$
Note that the intervention measure on  $A$  is equal to  $K_{S\setminus U}(\omega_0,A)$
$$
\mathbb {P} ^ {\mathrm {d o} (S \backslash U, \delta_ {\omega_ {0}})} (A) = \int \delta_ {\omega_ {0}} (d \omega_ {S \backslash U} ^ {\prime}) K _ {S \backslash U} (\omega^ {\prime}, A) = K _ {S \backslash U} (\omega_ {0}, A).
$$
Putting these together, we have
$$
K _ {S \cap U} ^ {\mathrm {d o} (S \backslash U, \delta_ {\omega_ {0}}, \text {h a r d})} (\omega_ {0}, A) \neq \mathbb {P} ^ {\mathrm {d o} (S \backslash U, \delta_ {\omega_ {0}})} (A),
$$
i.e. in the intervention causal space  $(\Omega, \mathcal{H}, \mathbb{P}^{\mathrm{do}(S \setminus U, \delta_{\omega_0})}, K_{S \cap U}^{\mathrm{do}(S \setminus U, \delta_{\omega_0}, \mathrm{hard})})$ ,  $\mathcal{H}_{S \cap U}$  has an active causal effect on  $A$ .
Lemma C.6. Let  $(\Omega, \mathcal{H}, \mathbb{P}, \mathbb{K}) = (\times_{t \in T} E_t, \otimes_{t \in T} \mathcal{E}_t, \mathbb{P}, \mathbb{K})$  be a causal space, and  $U, V \in \mathcal{P}(T)$ . For an event  $A \in \mathcal{H}$ , suppose that  $\mathcal{H}_U$  has no causal effect on  $A$  given  $\mathcal{H}_V$  (see Definition B.4). Then after an intervention on  $\mathcal{H}_V$  via any  $(\mathbb{Q}, \mathbb{L})$ ,  $\mathcal{H}_{U \setminus V}$  has no causal effect on  $A$ .
Proof. Take any probability measure  $\mathbb{Q}$  on  $(\Omega, \mathcal{H}_V)$  and any causal mechanism  $\mathbb{L}$  on  $(\Omega, \mathcal{H}_V, \mathbb{Q})$ . Then see that, for any  $S \in \mathcal{P}(T)$  and all  $\omega \in \Omega$ ,
$$
\begin{array}{l} K _ {S} ^ {\mathrm {d o} (V, \mathbb {Q}, \mathbb {L})} (\omega , A) = \int L _ {S \cap V} \left(\omega_ {S \cap V}, d \omega_ {V} ^ {\prime}\right) K _ {S \cup V} \left(\left(\omega_ {S \setminus V}, \omega_ {V} ^ {\prime}\right), A\right) \\ = \int L _ {S \cap V} \left(\omega_ {S \cap V}, d \omega_ {V} ^ {\prime}\right) K _ {(S \cup V) \backslash (U \backslash V)} \left(\left(\omega_ {S \backslash (U \cup V)}, \omega_ {V} ^ {\prime}\right), A\right) \\ = \int L _ {(S \backslash (U \backslash V)) \cap V} \left(\omega_ {(S \backslash (U \backslash V)) \cap V}, d \omega_ {V} ^ {\prime}\right) K _ {(S \backslash (U \backslash V)) \cup V} \left(\left(\omega_ {S \backslash (U \cup V)}, \omega_ {V} ^ {\prime}\right), A\right) \\ = K _ {S \backslash (U \backslash V)} ^ {\mathrm {d o} (V, \mathbb {Q}, \mathbb {L})} (\omega , A), \\ \end{array}
$$
where, in going from the first line to the second, we used the fact that  $\mathcal{H}_U$  has no causal effect on  $A$  given  $\mathcal{H}_V$ , and in going from the second line to the third, we used identities  $S \cap V = (S \setminus (U \setminus V)) \cap V$  and  $(S \cup V) \setminus (U \setminus V) = (S \setminus (U \setminus V)) \cup V$ . Since  $S \in \mathcal{P}(T)$  was arbitrary, we have that  $\mathcal{H}_{U \setminus V}$  has no causal effect on  $A$  in the intervention causal space.
Proof. Take any  $w_{1}, w_{\mathbb{Q}} \in W$  with  $w_{1} < w_{2}$ . Since  $\mathbb{K}$  respects time, we have that  $\mathcal{H}_{w_2 \times \tilde{T}}$  has no causal effect on  $\mathcal{H}_{w_1 \times \tilde{T}}$  in the original causal space. To show that  $\mathcal{H}_{w_2 \times \tilde{T}}$  has no causal effect on  $\mathcal{H}_{w_1 \times \tilde{T}}$  after a hard intervention on  $\mathcal{H}_U$  via  $\mathbb{Q}$ , take any  $S \in \mathcal{P}(T)$  and any event  $A \in \mathcal{H}_{w_1 \times \tilde{T}}$ . Then using Theorem C.3,
$$
\begin{array}{l} K _ {S} ^ {\mathrm {d o} (U, \mathbb {Q}, \mathrm {h a r d})} (\omega , A) \\ = \int \mathbb {Q} (d \omega^ {\prime}) K _ {S \cup U} ((\omega_ {S}, \omega_ {U \setminus S} ^ {\prime}), A) \\ = \int \mathbb {Q} (d \omega^ {\prime}) K _ {(S \cup U) \backslash \mathcal {H} _ {w _ {2} \times \tilde {T}}} ((\omega_ {S \backslash \mathcal {H} _ {w _ {2} \times \tilde {T}}}, \omega_ {U \backslash (S \cup \mathcal {H} _ {w _ {2} \times \tilde {T}})} ^ {\prime}), A) \\ = \int \mathbb {Q} (d \omega^ {\prime}) K _ {((S \cup U) \backslash \mathcal {H} _ {w _ {2} \times \hat {T}}) \cup (U \cap \mathcal {H} _ {w _ {2} \times \hat {T}})} ((\omega_ {S \backslash \mathcal {H} _ {w _ {2} \times \hat {T}}}, \omega_ {(U \backslash (S \cup \mathcal {H} _ {w _ {2} \times \hat {T}})) \cup (U \cap \mathcal {H} _ {w _ {2} \times \hat {T}})} ^ {\prime}), A) \\ = \int \mathbb {Q} (d \omega^ {\prime}) K _ {(S \backslash \mathcal {H} _ {w _ {2} \times \bar {T}}) \cup U} ((\omega_ {S \backslash \mathcal {H} _ {w _ {2} \times \bar {T}}}, \omega_ {U \backslash (S \backslash \mathcal {H} _ {w _ {2} \times \bar {T}})} ^ {\prime}), A) \\ = K _ {S \backslash \mathcal {H} _ {w _ {2} \times \tilde {T}}} ^ {\mathrm {d o} (U, \mathbb {Q}, \text {h a r d})} (\omega , A) \\ \end{array}
$$
where, from the second line to the third, we used the fact that  $\mathcal{H}_{w_2\times \tilde{T}}$  has no causal effect on  $A$ , from the third line to the fourth we used the fact that  $U\cap \mathcal{H}_{w_2\times \tilde{T}}$  has no causal effect on  $A$  (by Remark B.2(e)) and Remark B.2(g), and from the fourth line to the fifth, we used that  $((S\cup U)\backslash \mathcal{H}_{w_2\times \tilde{T}})\cup (U\cap \mathcal{H}_{w_2\times \tilde{T}}) = (S\setminus \mathcal{H}_{w_2\times \tilde{T}})\cup U$  and  $(U\setminus (S\cup \mathcal{H}_{w_2\times \tilde{T}}))\cup (U\cap \mathcal{H}_{w_2\times \tilde{T}}) =$ $U\setminus (S\setminus \mathcal{H}_{w_2\times \tilde{T}})$ . Since  $S\in \mathbb{P}(T)$  was arbitrary, we have that  $\mathcal{H}_{w_2\times \tilde{T}}$  has no causal effect on  $A$  (Definition B.1(i)). Since  $A\in \mathcal{H}_{w_1\times \tilde{T}}$  was arbitrary,  $\mathcal{H}_{w_2\times \tilde{T}}$  has no causal effect on  $\mathcal{H}_{w_1\times \tilde{T}}$ , and so  $\mathbb{K}^{\mathrm{do}(U,\mathbb{Q},\mathrm{hard})}$  respects time.