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*}

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*}

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 $Z:(\Omega,\mathcal F)\to(E,\mathcal E)$ 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.

Let $X\in L^1(\Omega,\mathcal F,\mathbb P)$. If $\mathcal G=\{\varnothing,\Omega\}$, then \begin{align*} \mathbb E[X\mid\mathcal G]=\mathbb E[X] \end{align*} $\mathbb P$-a.e. If $X$ is $\mathcal G$-measurable, then \begin{align*} \mathbb E[X\mid\mathcal G]=X \end{align*} $\mathbb P$-a.e.

Let $1\le p<\infty$, let $X\in L^p(\Omega,\mathcal F,\mathbb P)$, and let $\mathcal G\subset\mathcal F$ be a sub-$\sigma$-algebra. Then \begin{align*} \|\mathbb E[X\mid\mathcal G]\|_{L^p}\le\|X\|_{L^p}. \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.