# Independence under regular conditional probability

Let $$\{X_j\}_{j=1}^{n}$$ be a independent identically distributed random variables taking values in $$\mathbb{R}^d$$. We write $$\mu$$ for the distributions. We assume moreover that $$\mu$$ is absolutely continuous to the $$d$$-dim Lebesgue measure.

For $$x \in \mathbb{R}^d$$ and $$r>0$$, we denote by $$B_x(r) \subset \mathbb{R}^d$$ the closed ball centered at $$x$$ with radius $$r>0$$. Then For $$x \in \mathbb{R}^d$$ and $$A \in \mathcal{F}$$, we can consider $$P\left[A \mid X_1=x \right]$$, the regular conditional probability given $$X_1=x$$:

My question. In a published paper, the authors conclude that \begin{align*} &P\left[\bigcap_{j=2}^{n-1}\{X_{j} \notin B(X_n,|x-X_n|)\} \mid X_1=x \right]=P\left[\bigcap_{j=2}^{n-1}X_{j} \notin B(X_n,|x-X_n|) \right]. \end{align*}

I do not know the reason why this identity is valid... Why is this true? The left-hand side should be interpreted as $$P[A_x \mid X_1=x]$$. Here, $$A_x=\bigcap_{j=2}^{n-1}\{X_{j} \notin B(X_n,|x-X_n|)\} \in \mathcal{F}$$ is a measurable set. From the definition of the regular conditional probability, can we obtain the following identity?: for any measurable $$C$$, \begin{align*} &\int_{C}P\left[\bigcap_{j=2}^{n-1}X_{j} \notin B(X_n,|x-X_n|) \mid X_1=x \right]\,d\mu(x) =E\left[\bigcap_{j=2}^{n-1}X_{j} \notin B(X_n,|X_1-X_n|) ,\,X_1 \in C \right]. \end{align*} I feel like it's impossible because the event $$A_x$$ is dependent on $$x$$.

$$\newcommand\ov\overline\newcommand\R{\mathbb R}$$This is just an application of Tonelli's theorem. Indeed, let $$X:=X_1$$, $$Y:=(Y_2,\dots,Y_n)$$, $$Y_i:=X_i$$ for $$i\in\ov{2,n}$$, where $$\ov{k,l}:=[k,l]\cap\mathbb Z$$. Let \begin{align*} A&:=\{(x,y):=(x,y_2,\dots,y_n)\in(\R^d)^n \colon\\ &\qquad\qquad\qquad\forall j\in\ov{2,n-1}\ y_j\notin B_{y_n}(|x-y_n|)\} \\ &:=\{(x,y)\in(\R^d)^n \colon\forall j\in\ov{2,n-1}\ |y_j-y_n|>|x-y_n|\}, \end{align*} so that $$A$$ is a Borel set. Let
$$\begin{equation*} f:=1_A. \end{equation*}$$ Then for any Borel $$C\subseteq\R^d$$ \begin{align*} P(\{(X,Y)\in A\}\cap\{X\in C\})&=Ef(X,Y)1(X\in C\} \\ &=\int_{\R^d} P(X\in dx)\int_{(\R^d)^{n-1}} P(Y\in dy)f(x,y)1(x\in C) \\ &=\int_{\R^d} P(X\in dx)1(x\in C)\,Ef(x,Y) \\ &=\int_C P(X\in dx)\,P(Y\in A_x), \end{align*} where $$A_x:=\{y\in(\R^d)^{n-1}\colon(x,y)\in A\}$$; the second equality in the above display follows by Tonelli's theorem and the independence of $$X$$ and $$Y$$.
So, the regular conditional probability of the event $$\{(X,Y)\in A\}$$ given $$X=x$$ is $$P(Y\in A_x)$$, as desired.
More general: Let $$(\Omega,\mathcal{A},\mathbb{P})$$ a probability space and $$X \colon \Omega \to E$$ and $$Y \colon \Omega \to F$$ be independent random variables with values in Polish spaces $$E$$ and $$F$$ and $$\mathcal{B}(E)$$ be the Borel space of $$E$$, similarly for $$F$$. Then $$\mathcal{B}(E \times F) = \mathcal{B}(E) \otimes \mathcal{B}(F)$$. Let $$A \in \mathcal{B}(E) \otimes \mathcal{B}(F)$$ be arbitrary. Then for all $$B \in \mathcal{B}(E)$$ first by the theorem of Fubini and second by the definition of conditional expectation
$$\mathbb{P}((X,Y) \in A \cap B \times F) = \mathbb{P}^X \otimes \mathbb{P}^Y(A \cap B \times F) = \int_B \mathbb{P}^X(dx) \mathbb{P}^Y(A_x) = \int \mathbb{P}_B^X(dx) \mathbb{P}(A_x|X=x).$$ It follows that $$\mathbb{P}^Y(A_x) = \mathbb{P}(A_x|X=x)$$ for $$\mathbb{P}^X$$-a.e. $$x \in E$$.