Let $H$ be a [Hilbert space](/page/Hilbert%20Space). An orthonormal system in $H$ is maximal if and only if it is an [orthonormal basis](/page/Orthonormal%20Basis) of $H$.

Let $H$ be a [Hilbert space](/page/Hilbert%20Space) over $\mathbb{R}$ or $\mathbb{C}$, and let $(e_i)_{i \in I}$ be an [orthonormal basis](/page/Orthonormal%20Basis) of $H$, meaning that $(e_i)_{i \in I}$ is an orthonormal family and \begin{align*} \overline{\operatorname{span}\{e_i : i \in I\}} = H. \end{align*} Define $\ell^2(I)$ to be the Hilbert space of scalar families $a = (a_i)_{i \in I}$ such that \begin{align*} \|a\|_{\ell^2(I)}^2 := \sup\left\{\sum_{i \in F} |a_i|^2 : F \subset I \text{ finite}\right\} < \infty. \end{align*} Then the coordinate map \begin{align*} U: H &\to \ell^2(I) \\ x &\mapsto \bigl((x,e_i)_H\bigr)_{i \in I} \end{align*} is a linear isometric isomorphism. Its inverse is the map \begin{align*} V: \ell^2(I) &\to H \\ (a_i)_{i \in I} &\mapsto \lim_{F} \sum_{i \in F} a_i e_i, \end{align*}…

Let $(\Omega,\mathcal F)$ be a measurable space, and let $(E_i,\mathcal E_i)$ be measurable spaces for $i \in \{1,\dots,n\}$. A map \begin{align*} X: \Omega &\to E_1 \times \cdots \times E_n \end{align*} with coordinate maps $X_i: \Omega \to E_i$ is measurable from $(\Omega,\mathcal F)$ to $(E_1 \times \cdots \times E_n,\mathcal E_1 \otimes \cdots \otimes \mathcal E_n)$ iff each coordinate map $X_i$ is measurable.

Assuming the [axiom of choice](/page/Axiom%20of%20Choice), there exists a subset $V \subsetneq [0,1]$ that does not belong to the Lebesgue sigma-algebra on $\mathbb R$.

Let $(\Omega,\mathcal F)$ be a measurable space and let $(A_n)_{n=1}^{\infty}$ be a sequence of events. Then \begin{align*} \limsup_{n \to \infty} A_n \in \mathcal F, \qquad \liminf_{n \to \infty} A_n \in \mathcal F. \end{align*}

Let $\mathcal A \subset \mathcal P(\Omega)$. Then $\sigma(\mathcal A)$ is a sigma-algebra on $\Omega$, contains $\mathcal A$, and satisfies the following minimality property: if $\mathcal G$ is any sigma-algebra on $\Omega$ with $\mathcal A \subset \mathcal G$, then $\sigma(\mathcal A) \subset \mathcal G$.

Let $\Omega$ be a set. For each $i \in I$, let $(E_i,\mathcal E_i)$ be a measurable space and let $X_i: \Omega \to E_i$ be a function. Then $\sigma(X_i : i \in I)$ is the smallest sigma-algebra $\mathcal F$ on $\Omega$ for which every map $X_i: (\Omega,\mathcal F) \to (E_i,\mathcal E_i)$ is measurable.