Let \((\Omega, \mathcal E, \text{Pr})\) be a probability space, where \(\Omega\) is the space of random outcomes, \(\mathcal E\) the \(\sigma\)-algebra of measurable events, and \(P\) the probability measure.

Given a random variable \(T\colon \Omega \to \mathbb R\), define \(p_T\colon \Omega \to \left[0,1\right]\) as:

\[ p_T(\omega) = \text{Pr}(\{\omega'\in \Omega,円\vert,円 T(\omega')\geq T(\omega)\}) \]

Theorem. \(p_T\) is measurable and \(\text{Pr}(p_T\leq \alpha) \leq \alpha\) for all \(\alpha \in \left[0,1\right]\). Equality holds if and only if there exists a sequence \(\{\omega_n\}_{n\in \mathbb N}\) such that \(p_T(\omega_n) \leq \alpha\), and \(p_T(\omega _n)\to \alpha\) as \(n \to \infty\).

Proof. Let \(\alpha\in\left[0,1\right]\), and denote: \[ E_T(\omega) = \{\omega'\in \Omega,円\vert,円 T(\omega')\geq T(\omega)\}, \] so that \(p_T(\omega) = \text{Pr}(E_T(\omega))\).

Assume first that there exists \(\omega_\alpha \in p_T^{-1}(\alpha)\), that is to say \(\text{Pr}(E_T(\omega)) = \alpha\). We can show that:

\[ N_T(\omega_\alpha) = \{\omega \vert p_T(\omega) \leq \alpha\} \backslash E_T(\omega_\alpha) \] is a measurable, zero probability set, which proves the thesis for this particular case. As a matter of fact, for any \(\omega \in \Omega\), if \(p_T(\omega)\leq \alpha\) and \(T(\omega) <T(\omega_\alpha)\), then we must have:

\[ \text{Pr}(\{\omega'\in \Omega,円\vert,円 T(\omega_\alpha)>T(\omega')\geq T(\omega)\} ) = p_T(\omega) - \alpha=0. \] If \(t_* = \inf_{p_T(\omega)\leq \alpha}T(\omega)\) and \(\{a _n\}_{n \in \mathbb N}\) is a sequence in \(\Omega\) such that \(T(a _n)\to t_*\) as \(n\to \infty\), then:

\[ N_T(\omega _\alpha) \subseteq \cup _n \{\omega'\in \Omega,円\vert,円 T(\omega_\alpha)>T(\omega')\geq T(a_n)\}, \] the right hand side being a probability zero set.

If \(p_T^{-1}(\alpha)\) is empty, let \(\alpha^* = \sup _{p_T(\omega)\leq \alpha}p(\omega)\), and let \(\{b _n\}_{n\in \mathbb N}\) be a sequence in \(\Omega\) such that \(p_T(b_n)\to \alpha^*\) as \(n\to \infty\). Then:

\[ \{\omega \vert p_T(\omega) \leq \alpha\}= \{\omega \vert p_T(\omega) \leq \alpha^*\}= \cup _n \{\omega \vert p_T(\omega) \leq p_T(b_n)\}, \] so that, from the particular case proved earlier, we have:

\[ \text{Pr}(p_T \leq \alpha) = \lim _{n \to \infty} \text{Pr}(p_T \leq p_T(b_n)) \leq \lim _{n \to \infty} p_T(b_n) = \alpha ^* \leq \alpha, \] as was to be proved.