Showing $P[X_1<\cdots < X_n] = 1/n!$ when $X_i$ do not have densities. Suppose $(X_i)_i$ are independent and identically distributed from the same continuous distribution. If the distribution is absolutely continuous (i.e. the $X_i$ have density functions)  then showing $P[X_1<\cdots< X_n]= \frac{1}{n!}$ follows quickly from Fubini's theorem and a counting argument.
How can one show this fact without using density functions?
 A: Okay I figured it out myself. Firstly, let us show $P[X_i=X_j]=0$ for $i\neq j$. Let $F(t)$ denote the common cdf, by assumption $F$ is continuous. Partition $[0,1]$ into $n$ equal width subintervals, and let $A_{n,1},\ldots, A_{n,n}$ denote the preimages of these intervals under $F$. Then $P[A_{n,k}] = 1/n$ for all $n$. For all $n$ we have $$\{X_i=X_j\} \subseteq \cup_{k=1}^n\{X_i \in A_{n,k}, X_j \in A_{n,k}\}$$
and so
$$
P[X_i=X_j] \leq \sum_{k=1}^nP[X_i \in A_{n,k}, X_j \in A_{n,k}] = \sum_{k=1}^nP[X_i \in A_{n,k}]P[X_j \in A_{n,k}] = n \frac{1}{n^2} = \frac{1}{n}
$$
and taking the limit as $n \to \infty$ shows $P[X_i=X_j]=0$.
In light of the previous fact, we have $$\sum_{\sigma \in S_n} P[X_{\sigma(1)} < \cdots < X_{\sigma(n)}] = 1.$$
We then consider that for $N>n$ we may choose any $n$ distinct $A_{N,k_1},\ldots,A_{N,k_n}$ (where the $k_i$ are increasing) and then the event $$\{X_1 \in A_{N,k_1}, \ldots, X_n \in A_{N,k_n}\} \subseteq \{X_1 < \cdots < X_n\} $$
There are ${N \choose n}$ distinct choices for $k_1,\ldots,k_n$ and the corresponding events are disjoint and have probability $\frac{1}{N^n}$, so
$$
\frac{1}{N^n}{N \choose n} \leq P[X_1 < \cdots < X_n]
$$
and taking the limit as $N \to \infty$ gives
$$
\frac{1}{n!} \leq P[X_1 < \cdots < X_n].
$$
The same argument of course applies for any $\sigma \in S_n$, and summing them shows that none of the inequalities can be strict, else the sum would exceed $1$. Hence $ P[X_1 < \cdots < X_n]=\frac{1}{n!}$.
