Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, and let $X:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ be a real-valued random variable. Let $\mathcal L^1$ denote one-dimensional Lebesgue measure on $(\mathbb R,\mathcal B(\mathbb R))$. Then $X$ is absolutely continuous, meaning that its law $\mu_X:=\mathbb P\circ X^{-1}$ satisfies $\mu_X\ll\mathcal L^1$, if and only if there exists a nonnegative Borel measurable function $f_X:\mathbb R\to[0,\infty]$ such that \begin{align*} \mathbb P(X\in A)=\int_A f_X(x)\,d\mathcal L^1(x) \end{align*} for every $A\in\mathcal B(\mathbb R)$. The function $f_X$ is unique up to equality $\mathcal L^1$-a.e.

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, let $(E,\mathcal E)$ be a measurable space, and let $X:(\Omega,\mathcal F)\to(E,\mathcal E)$ be a [random variable](/page/Random%20Variable), meaning an $\mathcal F/\mathcal E$-measurable map. Define the distribution, or pushforward measure, $\mu_X:\mathcal E\to[0,1]$ by \begin{align*} \mu_X(A)=\mathbb P(X^{-1}(A)) \end{align*} for every $A\in\mathcal E$, where $X^{-1}(A)=\{\omega\in\Omega:X(\omega)\in A\}$. Then for every nonnegative $\mathcal E$-measurable function $g:E\to[0,\infty]$, \begin{align*} \int_\Omega g(X(\omega))\,d\mathbb P(\omega)=\int_E g(x)\,d\mu_X(x). \end{align*}

Let $G$ and $H$ be groups written multiplicatively, let $S \subset G$, and suppose $G = \langle S\rangle$, where $\langle S\rangle$ denotes the subgroup of $G$ generated by $S$. If $\varphi,\psi: G \to H$ are group homomorphisms such that $\varphi(s)=\psi(s)$ for every $s \in S$, then $\varphi=\psi$ as functions $G \to H$.

Let $(G,\cdot_G)$ and $(H,\cdot_H)$ be groups with identity elements $1_G$ and $1_H$, respectively. Let $\varphi:G\to H$ be a [group homomorphism](/page/Group%20Homomorphism), meaning that \begin{align*} \varphi(a\cdot_G b)=\varphi(a)\cdot_H\varphi(b) \end{align*} for all $a,b\in G$. Then \begin{align*} \varphi(1_G)&=1_H, \\ \varphi(g^{-1})&=\varphi(g)^{-1} \end{align*} for every $g\in G$, where $g^{-1}$ denotes the inverse of $g$ in $G$ and $\varphi(g)^{-1}$ denotes the inverse of $\varphi(g)$ in $H$.

Let $G$ be a group with centre \begin{align*} Z(G)=\{g\in G: gx=xg \text{ for every } x\in G\}. \end{align*} Let $\operatorname{Aut}(G)$ denote the group of automorphisms of $G$, and let $\operatorname{Inn}(G)\le \operatorname{Aut}(G)$ be the subgroup consisting of automorphisms of the form $x\mapsto gxg^{-1}$ for some $g\in G$. Then \begin{align*} G/Z(G)\cong \operatorname{Inn}(G). \end{align*}

Let $G$ be a group with multiplication written multiplicatively. For each $a \in G$, define the conjugation map \begin{align*} \iota_a:G&\to G\\ x&\mapsto axa^{-1}. \end{align*} Then the function \begin{align*} \Phi:G&\to \operatorname{Aut}(G)\\ a&\mapsto \iota_a \end{align*} is a [group homomorphism](/page/Group%20Homomorphism), where $\operatorname{Aut}(G)$ is the automorphism group of $G$ under composition.

Let $(G,\cdot)$ be a group, and let $\operatorname{Aut}(G)$ denote the set of all group automorphisms $\varphi: G \to G$. Then $\operatorname{Aut}(G)$ is a group under function composition.

Let $(X,\tau_X)$ be a first-countable [topological space](/page/Topological%20Space), let $(Y,\tau_Y)$ be a topological space, and let $f:X\to Y$ be a function. Then $f$ is continuous if and only if, for every point $x\in X$ and every sequence $(x_k)_{k\in\mathbb N}$ in $X$, convergence $x_k\to x$ in $(X,\tau_X)$ implies convergence $f(x_k)\to f(x)$ in $(Y,\tau_Y)$.

Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be topological spaces, let $f:X\to Y$ be continuous, and let $(x_k)_{k\in\mathbb{N}}$ be a sequence in $X$. If $x\in X$ and $x_k\to x$ in $(X,\tau_X)$, then $f(x_k)\to f(x)$ in $(Y,\tau_Y)$.

Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be topological spaces. Let $A \subset Y$, and equip $A$ with the [subspace topology](/page/Subspace%20Topology) \begin{align*} \tau_A := \{A \cap V : V \in \tau_Y\}. \end{align*} Let $f: X \to A$ be a function, and let $i: A \to Y$ be the inclusion map, defined by $i(a)=a$ for every $a \in A$. Then $f: (X,\tau_X) \to (A,\tau_A)$ is continuous if and only if $i \circ f: (X,\tau_X) \to (Y,\tau_Y)$ is continuous.