Let \(\mu \) be a finite measure, let \(X:\Omega \to \mathbb {R}\) be measurable and supported almost everywhere on a finite set \(s\subseteq \mathbb {R}\), and let \(f:\mathbb {R}\to \mathbb {R}\) be arbitrary. Then \(\int f(X)\, d\mu =\sum _{x\in s}\mu (X=x)f(x)\).

Let \(\mu \) be a finite measure, let \(X\) be measurable and supported almost everywhere on a finite \(s\subseteq \mathbb {R}\), let \(q\gt 0\), and let \(a,b\in \mathbb {R}\). Then \(\| a+bX\| _q^2=(\sum _{x\in s}\mu (X=x)|a+bx|^q)^{2/q}\).

Let \(X\) be a measurable real random variable on a probability space and let \(x\ne y\). If \(\Pr [X=x]+\Pr [X=y]=1\), then \(X\in \{ x,y\} \) almost surely.

Let \(X\) be measurable on a probability space, let \(x\ne y\) and \(0\le \lambda \le 1\), and suppose \(\Pr [X=x]=\lambda \) and \(\Pr [X=y]=1-\lambda \). For \(q\gt 0\) and \(a,b\in \mathbb {R}\), \(\| a+bX\| _q^2=(\lambda |a+bx|^q+(1-\lambda )|a+by|^q)^{2/q}\).

Let \(X\) be measurable on a probability space and let \(0\le \lambda \le 1\). If \(\Pr [X=1-\lambda ]=\lambda \) and \(\Pr [X=-\lambda ]=1-\lambda \), then \(\mathbb {E}X=0\) and \(\| X\| _2^2=\lambda (1-\lambda )\).

Let \(X\) be measurable on a probability space and supported almost surely on a finite set \(s\subseteq \mathbb {R}\). If \(\lambda \ge 0\) and \(\Pr [X=x]\ge \lambda \) for every \(x\in s\), then the real atom weights \(w_x=\Pr [X=x]\) satisfy \(\sum _{x\in s}w_x=1\) and \(w_x\ge \lambda \) for every \(x\in s\).