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 $x$.