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 $x$.



Independence under regular conditional probability
$\newcommand\ov\overline\newcommand\R{\mathbb R}$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](https://en.wikipedia.org/wiki/Regular_conditional_probability) of the event $\{(X,Y)\in A\}$ given $X=x$ is $P(Y\in A\_x)$, as desired.


