[proofplan] We first prove that ties among the sample values occur with probability zero. On the no-tie event, the sample points have exactly one strict ordering, and each strict ordering determines exactly one rank vector. Independence and identical distribution make all strict ordering events have the same probability, while these events are disjoint and cover the no-tie event; since there are $n!$ of them, each has probability $1/n!$. [/proofplan]

[step:Show that ties occur with probability zero]Let $(\Omega,\mathcal F,\mathbb P)$ be the probability space from the theorem statement. Let $\mu$ denote the common distribution of the random variables $X_1,\dots,X_n$, so $\mu(A)=\mathbb{P}(X_1 \in A)$ for every $A \in \mathcal{B}(\mathbb{R})$. Since the common distribution is continuous, we use the standard atomlessness consequence that $\mu(\{x\})=0$ for every $x\in\mathbb{R}$. For distinct indices $i,j \in \{1,\dots,n\}$, independence of $X_i$ and $X_j$ gives the identity \begin{align*} \mathbb{P}(X_i = X_j) = (\mu \otimes \mu)\bigl(\{(x,y)\in \mathbb{R}^2 : x=y\}\bigr). \end{align*} Let $D := \{(x,y)\in \mathbb{R}^2 : x=y\}$ denote the diagonal. Since the indicator function $\mathbb{1}_D: \mathbb{R}^2 \to \{0,1\}$ is non-negative and Borel measurable, Tonelli's Theorem applies to the product measure $\mu \otimes \mu$. For each $x \in \mathbb{R}$, the vertical section $D_x := \{y \in \mathbb{R} : (x,y) \in D\}$ is the singleton $\{x\}$, and atomlessness of $\mu$ gives $\mu(D_x)=\mu(\{x\})=0$. Therefore \begin{align*} (\mu \otimes \mu)(D) = \int_{\mathbb{R}} \mu(D_x)\,d\mu(x) = 0. \end{align*} Hence \begin{align*} \mathbb{P}(X_i = X_j)=0. \end{align*} Define the no-tie event \begin{align*} E := \{\omega \in \Omega : X_i(\omega) \neq X_j(\omega)\text{ for all }i \neq j\}. \end{align*} By the union bound over the finite set of unordered pairs, \begin{align*} \mathbb{P}(E^c) \leq \sum_{1 \leq i < j \leq n}\mathbb{P}(X_i=X_j) =0. \end{align*} Therefore $\mathbb{P}(E)=1$.[/step]

[guided]The only obstruction to the ranks forming a permutation is the possibility of ties. We remove that obstruction first. Let $(\Omega,\mathcal F,\mathbb P)$ be the probability space from the theorem statement. Let $\mu$ be the common distribution of the random variables $X_1,\dots,X_n$. Thus for every Borel set $A \subset \mathbb{R}$, \begin{align*} \mu(A)=\mathbb{P}(X_1 \in A). \end{align*} Because the variables are identically distributed, this is also the distribution of each $X_i$. Because the common distribution is continuous, it is atomless: for every $x\in\mathbb{R}$, the singleton $\{x\}$ has $\mu$-measure zero. Fix two distinct indices $i,j \in \{1,\dots,n\}$. Since $X_i$ and $X_j$ are independent and have common law $\mu$, the random vector $(X_i,X_j)$ has distribution $\mu\otimes\mu$ on $\mathbb{R}^2$. Therefore \begin{align*} \mathbb{P}(X_i=X_j) = (\mu \otimes \mu)\bigl(\{(x,y)\in \mathbb{R}^2:x=y\}\bigr). \end{align*} Let $D := \{(x,y)\in \mathbb{R}^2 : x=y\}$ denote the diagonal. The set $D$ is closed in $\mathbb{R}^2$, hence Borel measurable, so its indicator function $\mathbb{1}_D: \mathbb{R}^2 \to \{0,1\}$ is Borel measurable. Since $\mathbb{1}_D$ is non-negative, Tonelli's Theorem applies to the product measure $\mu \otimes \mu$ and gives \begin{align*} (\mu \otimes \mu)(D) = \int_{\mathbb{R}} \left(\int_{\mathbb{R}} \mathbb{1}_D(x,y)\,d\mu(y)\right)d\mu(x). \end{align*} For a fixed $x \in \mathbb{R}$, the inner integral is the $\mu$-measure of the vertical section \begin{align*} D_x := \{y \in \mathbb{R} : (x,y) \in D\}=\{x\}. \end{align*} Because the common distribution is continuous, the measure $\mu$ has no atoms, so $\mu(\{x\})=0$ for every $x \in \mathbb{R}$. Hence \begin{align*} (\mu \otimes \mu)(D) = \int_{\mathbb{R}} 0\,d\mu(x) = 0. \end{align*} Therefore \begin{align*} \mathbb{P}(X_i=X_j)=0. \end{align*} Now define the event on which no two sample values are equal: \begin{align*} E := \{\omega \in \Omega : X_i(\omega) \neq X_j(\omega)\text{ for all }i \neq j\}. \end{align*} Its complement is the event that at least one pair ties. Since there are only finitely many unordered pairs, the union bound gives \begin{align*} \mathbb{P}(E^c) \leq \sum_{1 \leq i < j \leq n}\mathbb{P}(X_i=X_j) =0. \end{align*} Thus $\mathbb{P}(E^c)=0$, equivalently $\mathbb{P}(E)=1$. This is the point at which the continuity, equivalently atomlessness, of the common distribution is used.[/guided]

[step:Partition the no-tie event into strict ordering events] Let $S_n$ denote the set of all permutations of $\{1,\dots,n\}$. For each $\pi \in S_n$, define the strict ordering event \begin{align*} A_\pi := \{\omega \in \Omega : X_{\pi(1)}(\omega) < X_{\pi(2)}(\omega) < \cdots < X_{\pi(n)}(\omega)\}. \end{align*} On $E$, exactly one of the events $A_\pi$ occurs, because the $n$ distinct [real numbers](/page/Real%20Numbers) $X_1(\omega),\dots,X_n(\omega)$ have a unique increasing order. Therefore the events $(A_\pi)_{\pi \in S_n}$ are pairwise disjoint and satisfy \begin{align*} E = \bigcup_{\pi \in S_n} A_\pi. \end{align*} [/step]

[step:Use exchangeability to give every strict ordering the same probability] Fix $\pi,\rho \in S_n$. Since $X_1,\dots,X_n$ are independent and identically distributed, the random vectors \begin{align*} (X_{\pi(1)},\dots,X_{\pi(n)}) \quad\text{and}\quad (X_{\rho(1)},\dots,X_{\rho(n)}) \end{align*} have the same distribution on $\mathbb{R}^n$. Applying this equality of distributions to the Borel set \begin{align*} B := \{(x_1,\dots,x_n)\in \mathbb{R}^n : x_1 < x_2 < \cdots < x_n\}, \end{align*} we obtain \begin{align*} \mathbb{P}(A_\pi) = \mathbb{P}(A_\rho). \end{align*} Thus all strict ordering events have one common probability. Denote this common probability by $q \in [0,1]$. [/step]

[step:Compute the common ordering probability] Since the events $(A_\pi)_{\pi \in S_n}$ are pairwise disjoint, cover $E$, and $\mathbb{P}(E)=1$, finite additivity gives \begin{align*} 1 = \mathbb{P}(E) = \sum_{\pi \in S_n}\mathbb{P}(A_\pi) = \sum_{\pi \in S_n} q = n!q. \end{align*} Hence \begin{align*} q=\frac{1}{n!}. \end{align*} Therefore \begin{align*} \mathbb{P}(A_\pi)=\frac{1}{n!} \end{align*} for every $\pi \in S_n$. [/step]

[step:Translate strict orderings into rank vectors] For $i\in\{1,\dots,n\}$, the rank $R_i$ is the increasing rank of $X_i$ among $X_1,\dots,X_n$, with rank $1$ assigned to the smallest observation. Let $\sigma \in S_n$ be a rank vector. Define the map $\pi_\sigma:\{1,\dots,n\}\to\{1,\dots,n\}$ by \begin{align*} \pi_\sigma(k) := \sigma^{-1}(k), \qquad k \in \{1,\dots,n\}. \end{align*} Since $\sigma$ is a permutation, $\pi_\sigma\in S_n$. On the no-tie event $E$, the condition \begin{align*} (R_1,\dots,R_n) = (\sigma(1),\dots,\sigma(n)) \end{align*} is equivalent to \begin{align*} X_{\pi_\sigma(1)} < X_{\pi_\sigma(2)} < \cdots < X_{\pi_\sigma(n)}. \end{align*} Thus the two events differ at most by the null set $E^c$, and so \begin{align*} \mathbb{P}\bigl((R_1,\dots,R_n) = (\sigma(1),\dots,\sigma(n))\bigr) = \mathbb{P}(A_{\pi_\sigma}) = \frac{1}{n!}. \end{align*} Since this holds for every $\sigma \in S_n$, the rank vector $(R_1,\dots,R_n)$ is uniformly distributed over all permutations of $(1,\dots,n)$. [/step]