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*}

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*}

The collection of Lebesgue measurable subsets of $\mathbb{R}^n$ is a $\sigma$-algebra containing all open subsets of $\mathbb{R}^n$. The measure $\mathcal{L}^n$ is countably additive on this $\sigma$-algebra. If $(E_k)_{k=1}^\infty$ are pairwise disjoint Lebesgue measurable sets, then \begin{align*} \mathcal{L}^n\left(\bigcup_{k=1}^\infty E_k\right) = \sum_{k=1}^\infty \mathcal{L}^n(E_k). \end{align*} Moreover, $\mathcal{L}^n(\varnothing)=0$ and $\mathcal{L}^n(E) \le \mathcal{L}^n(F)$ whenever $E \subset F$ are Lebesgue measurable.

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*}

Let $X,Y\in L^1(\Omega,\mathcal F,\mathbb P)$, let $a,b\in\mathbb R$, and let $\mathcal G\subset\mathcal F$ be a sub-$\sigma$-algebra. Then \begin{align*} \mathbb E[aX+bY\mid\mathcal G]=a\mathbb E[X\mid\mathcal G]+b\mathbb E[Y\mid\mathcal G] \end{align*} $\mathbb P$-a.e. If $X\ge0$ $\mathbb P$-a.e., then \begin{align*} \mathbb E[X\mid\mathcal G]\ge0 \end{align*} $\mathbb P$-a.e.

Let $Z:(\Omega,\mathcal F)\to(E,\mathcal E)$ be a random variable taking values in a standard Borel space, and let $Y:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ be a $\sigma(Z)$-measurable random variable. Then there exists a measurable function $g:E\to\mathbb R$ such that \begin{align*} Y=g(Z) \end{align*} $\mathbb P$-a.e.