A nearest neighbor conditional probability problem 
Let $(\mathcal{X},d)$ a metric space with its Borel $\sigma$-algebra $\mathcal{F}_{\mathcal{X}}$.
Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space.
Let $m\in\mathbb{N}$ with $m\ge2$ and $$X,X_1,...,X_m:(\Omega,\mathcal{F})\to(\mathcal{X},\mathcal{F}_{\mathcal{X}})$$ be $\mathbb{P}$-i.i.d. random variables.

Let $\sigma^1:[0,+\infty)^m\to\{1,...,m\}$ be a measurable function such that
$$\forall r_1,...,r_m\ge0, \sigma^1(r_1,...,r_m)\in \operatorname{argmin}_{k\in\{1,...,m\}} r_k$$
and $\sigma^2:[0,+\infty)^m\to\{1,...,m\}$ be a measurable function such that
$$\forall r_1,...,r_m\ge0, \sigma^2(r_1,...,r_m)\in \operatorname{argmin}_{k\in\{1,...,m\}\backslash \{\sigma^1(r_1,...,r_m)\}} r_k.$$

Let $x\in\mathcal{X}$.

Define $$\pi^1:\mathcal{X}^m\to\{1,...,m\}, (x_1,...,x_m)\mapsto\sigma^1(d(x,x_1),...,d(x,x_m))$$
and
$$\pi^2:\mathcal{X}^m\to\{1,...,m\}, (x_1,...,x_m)\mapsto\sigma^2(d(x,x_1),...,d(x,x_m)).$$
Define
$$X^1:\Omega\to\mathcal{X}, \omega \mapsto X_{\pi^1(X_1(\omega),...,X_m(\omega))},\\
X^2:\Omega\to\mathcal{X}, \omega \mapsto X_{\pi^2(X_1(\omega),...,X_m(\omega))},\\
W:\Omega\to[0,+\infty), \omega \mapsto d(x,X^2(\omega))$$

Intuitively $X^1$ and $X^2$ are respectively the first and the second random variables chosen from $X_1,...,X_m$ that are closer to $x$, and $W$ is the distance of $X^2$ from $x$.

Is it true that the distribution of $X^1$ given $W$ is equal to the distribution of $X$ given that the distance of $X$ from $x$ is less or equal then $W$? I.e.

is it true that
$$\forall A\in\mathcal{F}_{\mathcal{X}}, \mathbb{P}(X^1\in A | W) = \mathbb{P}(X\in A | d(x,X)\le W)?$$

Intuitively, it seems obvious: the closer random variable has to belong to the closed ball centered in $x$ of radius $W$ and, since the other random variables are banned from this ball, we have only one chance distributed as $\mathbb{P}_X$ bounded to this ball to hit $A$.
However I'm a bit in trouble trying to formalize this argument...
Any help?
 A: It seems that the claim is false in general since problems arise when there's a non-null probability of hitting a shell. What follows is a counter-example (at least if I haven't made any mistake).
Let $m=2$ and define
\begin{equation*}
\sigma^1:[0,+\infty)^2\to\{1,2\}, (r_1,r_2)\mapsto \min\left({\operatorname{argmin}_{k\in\{1,2\}}}r_k\right)
\end{equation*}
so that
\begin{equation*}
\forall r_1,r_2\ge 0, \sigma ^1(r_1,r_2)=
\begin{cases}
1, &\text{if $r_1 \le r_2$;}\\
2, &\text{otherwise;}\\
\end{cases}
\end{equation*}
and
\begin{equation*}
\forall r_1,r_2\ge 0, \sigma ^2(r_1,r_2)=
\begin{cases}
2, &\text{if $r_1 \le r_2$;}\\
1, &\text{otherwise.}\\
\end{cases}
\end{equation*}
Define $\mathcal{X}=\{0,1\}$ and let $d$ be the discrete metric on $\mathcal{X}$.
Let $p_0\in (0,1)$ and let $\mathbb{P}_X$ be the unique probability measure on $2^{\{0,1\}}$ such that $\mathbb{P}_{X}(\{0\})=p_0$. Let $x=0$ and $A=\{x\}$.
Let's show that
\begin{equation*}
\mathbb{P}\left(X^1 \in A | W=1\right) \neq \mathbb{P}\left(X \in A | d(x,X)\le 1\right)
\end{equation*}
First, notice that
\begin{equation*}
\mathbb{P}\left(X \in A | d(x,X)\le 1\right) = \mathbb{P}\left(X \in A\right) = \mathbb{P}(X=0) = p_0.
\end{equation*}
On the other hand
\begin{equation*}
\{W=1\} = \{X_2 = 1\} \cup \left(\{X_1 = 1\}\cap \{X_2 = 0\}\right)
\end{equation*}
and
\begin{align*}
\{X^1\in A\} \cap \{W=1\} &= \left(\{X^1=0\}\cap\{X_2 = 1\}\right) \cup \left(\{X^1=0\}\cap\{X_1 = 1\}\cap \{X_2 = 0\}\right)\\
&= \left(\{X_1=0\}\cap\{X_2 = 1\}\right) \cup \left(\{X_1 = 1\}\cap \{X_2 = 0\}\right),\\
\end{align*}
so
\begin{align*}
\mathbb{P}\left(X^1\in A | W=1\right) &= \frac{\mathbb{P}\left(\{X^1\in A\}\cap \{W=1\}\right)}{\mathbb{P}\left(\{W=1\}\right)}\\
              &= \frac{\mathbb{P}\left(\left(\{X_1=0\}\cap\{X_2 = 1\}\right) \cup \left(\{X_1 = 1\}\cap \{X_2 = 0\}\right)\right)}{\mathbb{P}\left(\{X_2 = 1\} \cup \left(\{X_1 = 1\}\cap \{X_2 = 0\}\right)\right)}\\
              &= \frac{2 p_0(1-p_0)}{1-p_0+p_0(1-p_0)}= \frac{2 p_0(1-p_0)}{1-p_0^2}\\
              &= \frac{2 p_0}{1+p_0}\neq p_0=\mathbb{P}\left(X \in A | d(x,X)\le 1\right).
\end{align*}
