$\mathbb{P}(\{X_n > X_0\} \cap \bigcap_{i=1}^{n-1}\{X_i \leq X_0\}) = \int_{0}^{1} \mathbb{P}(\{X_n > t\} \cap \bigcap_{i=1}^{n-1}\{X_i \le...$ Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, $X_1, ..., X_n$ some random variables from $\Omega$ to $[0, 1]$ independent and follow the same uniform law on $[0, 1]$.
Then, $\mathbb{P}(\{X_n > X_0\} \cap \bigcap_{i=1}^{n-1}\{X_i \leq X_0\}) = \int_{0}^{1} \mathbb{P}(\{X_n > t\} \cap \bigcap_{i=1}^{n-1}\{X_i \leq t\}) \mathrm{d} t$.
why this equality is true?
If the value of $X_0$ is given, I know calculate this probability by independence of $X_i$, but $\mathbb{P}$ is only $\sigma$-additive, so we can't sum by all $t\in[0, 1]$. I tried to make this more rigourous, so I  applied Fubini's theorem to the function $$(w, t)\in\Omega\times\mathbb{R} \mapsto 1_{\{X_n > X_0\}}(w)\prod_{i=1}^{n} 1_{\{X_i \leq X_0\}}(w)1_{\{X_0 = t\}}(w)$$, but it seems that dosen't work.
 A: You must calculate $$\int_A f(x_0, \dots,x_n) \mathrm{d}x_0\dots\mathrm{d}x_n$$
where $A= \{(x_0, \dots, x_n) \in [0,1]^n: \, x_n>x_0, \, x_i < x_0, \, i=1,\dots, n-1\}$ and $f$ is the joint density of the $X_i$.
You can observe that $A=[0,1] \times \{(x_1, \dots, x_n) \in [0,1]^{n-1}: \, x_n>x_0, \, x_i < x_0, \, i=1,\dots, n-1\}$, which I will write $A = [0,1] \times B$ for simplicity, and that the joint density can be written as the product of marginal densities due to the independece of the r.v.'s.
Now you are ready to apply Fubini's theorem:
$$\int_A f(x_0, \dots,x_n) \mathrm{d}x_0\dots\mathrm{d}x_n =\int_0^1\Bigg(\int_B f_{X_1,\dots,X_n}(x_1, \dots,x_n) \mathrm{d}x_1\dots\mathrm{d}x_n\Bigg)f_{X_0}(x_0)\mathrm{d}x_0 = \int_0^1\mathbb{P}\bigg(\{X_n > x_0\} \cap \bigcap_{i=1}^{n-1}\{X_i<x_0\}\bigg)\mathrm{d}x_0$$
where I've expressed the inner integral as a probability and I've used the fact that the uniform density on $[0,1]$ is $f_{X_0}(x_0)=1$.
A: Denote $A = \{(t_0, ..., t_n) \in [0, 1]^{n} \mid t_n>t_0, t_0\geq t_{n-1}, ..., t_0 \geq t_1\}$ and $h = x \in [0, 1]^{n} \mapsto 1_{A}(x)$.
We have $1_{\{X_n > X_0\}}(w)\prod_{i=1}^{n} 1_{\{X_i \leq X_0\}}(w) = h \circ (X_0, ..., X_n)(w)$.
So
$$\mathbb{P}(\{X_n > X_0\} \cap \bigcap_{i=1}^{n-1}\{X_i \leq X_0\}) = \int_{\Omega} h \circ (X_0, ..., X_n)(w) ~\mathrm{d}\mathbb{P}(w) = \int_{[0, 1]^{n}} h ~\mathrm{d} \mathbb{P}_{(X_0, ..., X_n)} = \int_{[0, 1]^{n}} h ~\mathrm{d} \mathbb{P}_{X_0} \otimes \cdots \otimes \mathrm{d} \mathbb{P}_{X_n} = \int_{[0, 1]^{n}} h ~\mathrm{d} \lambda \otimes \mathrm{d}\lambda^{\otimes n-1} = \int_{[0, 1]}\Big(\int_{[0, 1]^{n-1}} h(t_0, ..., t_n) ~\mathrm{d}\lambda^{\otimes n-1}(t_1, ..., t_{n-1})\Big)~\mathrm{d} \lambda(t_0) = \int_{[0, 1]} (1-t_0)t_{0}^{n-1} ~\mathrm{d} \lambda(t_0)$$
