Find the Lebesgue measure of simplex $\{(x_1, x_2, ... ,x_d)\in \mathbb R^d : 0 < x_1 < x_2 < ... < x_d < 1\}$

Question

Find the measure of simplex $\{(x_1, x_2, ... ,x_d)\in \mathbb R^d : 0 < x_1 < x_2 < ... < x_d < 1\}$.

How should I start?

Answer 1

The unit cube $[0,1]^d$ has measure $1$, and can be decomposed into $d!$ simplices (up to sets of measure zero) according to the $d!$ ways of ordering the coordinates. Hence the measure of each simplex is $\frac{1}{d!}$.

Or you can just compute: the measure is $$ \int_0^1\int_0^{x_d}\cdots\int_0^{x_2}\;dx_1\cdots dx_d=\int_0^1\int_0^{x_d}\cdots \int_0^{x_3}x_2\;dx_2\cdots dx_d=\cdots =\int_0^1\frac{x_d^{d-1}}{(d-1)!}\;dx_d=\frac{1}{d!} $$

Answer 2

Hint: Prove inductively that the set $\{x \mid 0 < x_1 < \ldots < x_n < t\}$ has Lebesgue measure $\frac{t^n}{n!}$

Answer 3

Alternatively, you can ask,

What is the probability that a random sequence of $n$ numbers between $0$ and $1$ is sorted?

and see that the answer is $\frac{1}{n!}$.

You can make this more rigorous by taking your set $S$ and defining, for each permutation $\sigma$ of $\{1,2,\dots,n\}$ a set: $$S_{\sigma}=(t_{\sigma 1},\dots,t_{\sigma n})\mid (t_1,\dots,t_n)\in S\}$$

These sets are disjoint, and all have the same measure. The subset of elements $[0,1]^n$ that are not in some $S_{\sigma}$ is the set $D=\{(t_1,\dots,t_n)\in[0,1]^n\mid \exists i\neq j: t_i=t_j\}.$ Show that $D$ has measure zero.