Let $V$ be a finite-dimensional inner product space and let $U\in \mathcal{L}(V)$. The following are equivalent: \begin{align*} &U\text{ is unitary},\\ &U^*U=I,\\ &\|Uv\|_V=\|v\|_V\text{ for all }v\in V. \end{align*}

Let $V$ be a finite-dimensional inner product space and let $W\le V$ be a subspace. For every $v\in V$, there exist unique vectors $w\in W$ and $z\in W^\perp$ such that \begin{align*} v=w+z. \end{align*} The vector $w$ is the unique minimizer of $\|v-y\|_V$ over all $y\in W$.

If $E \subset \mathbb{R}^n$ is Lebesgue measurable and $a \in \mathbb{R}^n$, then $E+a$ is Lebesgue measurable and \begin{align*} \mathcal{L}^n(E+a)=\mathcal{L}^n(E). \end{align*}

If $N \subset \mathbb{R}^n$ is a Lebesgue null set and $A \subset N$, then $A$ is Lebesgue measurable and \begin{align*} \mathcal{L}^n(A)=0. \end{align*}

For every Lebesgue measurable set $E \subset \mathbb{R}^n$, [Lebesgue measure](/page/Lebesgue%20Measure) is outer regular on $E$. If $\mathcal{L}^n(E)<\infty$, then Lebesgue measure is inner regular on $E$. Moreover, if $\mathcal{L}^n(E)<\infty$, then for every $\varepsilon>0$ there are an [open set](/page/Open%20Set) $U \subset \mathbb{R}^n$ and a compact set $K \subset \mathbb{R}^n$ such that \begin{align*} K \subset E \subset U, \qquad \mathcal{L}^n(U \setminus K) < \varepsilon. \end{align*}

Let $E \subset \mathbb{R}^n$ be Lebesgue measurable. Let $f: E \to [0,\infty]$ and $g: E \to [0,\infty]$ be Lebesgue [measurable functions](/page/Measurable%20Functions) such that $f=g$ for $\mathcal{L}^n$-a.e. $x \in E$. Then \begin{align*} \int_E f\, d\mathcal{L}^n = \int_E g\, d\mathcal{L}^n. \end{align*}