Measure Theory (I): $\sigma$-algebras

11 Sep 2018

• measure theory

The central concept of measure theory is that of a measure on a set $\Omega$, which is a way to quantify the “size” of its subsets. However, it is usually impossible to assign such a “size” to every subset of $\Omega$ in a way consistent with the axioms of a measure. In this post, we define the notion of a $\sigma$-algebra, which we think of as the collection of all “measurable” subsets of $\Omega$.

Preliminaries

Let $\Omega$ be a set. The power set of $\Omega$, denoted $\mc P(\Omega)$, is the collection of all subsets of $\Omega$.

A $\sigma$-algebra (or sigma-algebra) on $\Omega$ is a set $\Sigma\subseteq\mc P(\Omega)$, satisfying:

• $\emptyset\in\Sigma$;
• $A\in\Sigma\implies\Omega\backslash A\in\Sigma$;
• $A_1,A_2,\ldots\in\Sigma\implies\bigcup_{k=1}^\infty A_k\in\Sigma$.

In other words, a $\sigma$-algebra is a collection of subsets of $\Omega$ containing $\emptyset$, and is closed under complement and countable union. Note that closure under countable union also implies closure under finite union, by taking $A_k=\emptyset$ for all $k>N$.

Here are some examples of $\sigma$-algebras.

• Clearly, $\mc P(\Omega)$ is a $\sigma$-algebra on $\Omega$.
• By the axioms, any $\sigma$-algebra on $\Omega$ must contain $\emptyset$ and $\Omega\backslash\emptyset=\Omega$. Conversely, we can check that $\{\emptyset,\Omega\}$ is a $\sigma$-algebra on $\Omega$.
• Let $A\subseteq\Omega$. Then $\{\emptyset,A,\Omega\backslash A,\Omega\}$ is a $\sigma$-algebra on $\Omega$.

Example: Let $\lvert\Omega\rvert=4$, and let

$\Sigma=\{A\subseteq\Omega\,:\,|A|=0,\,2,\,\text{or }4\}.$

Then $\Sigma$ contains $\emptyset$, and is closed under complement and countable disjoint union. However, the union of two 2-element subsets of $\Omega$ can contain 3 elements, so $\Sigma$ is not a $\sigma$-algebra on $\Omega$.

Now we prove some elementary consequences of the axioms.

Proposition: For any $\sigma$-algebra $\Sigma$ on a set $\Omega$, we have:

• $A,B\in\Sigma\implies A\backslash B\in\Sigma$;
• $A_1,A_2,\ldots\in\Sigma\implies\bigcap_{k=1}^\infty A_k\in\Sigma$.

Proof: Note that

$\begin{aligned} A\backslash B&=A\cap(\Omega\backslash B),\\ \bigcap_{k=1}^\infty A_k&=\Omega\backslash\bigcup_{k=1}^\infty(\Omega\backslash A_k).\qed \end{aligned}$

In other words, $\sigma$-algebras are closed under countable intersection, and thus finite intersection (by taking $A_k=\Omega$ for $k>N$).

Proposition: If $\{\Sigma_\alpha\}_ {\alpha\in I}$ are $\sigma$-algebras on $\Omega$, then so is $\bigcap_{\alpha\in I}\Sigma_\alpha$.

Proof: Check that each of the axioms of a $\sigma$-algebra carry over to the intersection. $\qed$

For any collection $\mc S\subseteq\mc P(\Omega)$ of subsets of $\Omega$, we define the $\sigma$-algebra generated by $\mc S$, denoted by $\sigma(\mc S)$, to be the intersection of all $\sigma$-algebras on $\Omega$ containing $\mc S$.

Proposition: For any $\mc S\subseteq\mc P(\Omega)$, $\sigma(\mc S)$ is the smallest $\sigma$-algebra on $\Omega$ containing $\mc S$.

Proof: By the previous proposition, $\sigma(\mc S)$ is a $\sigma$-algebra. By definition, any $\sigma$-algebra containing $\mc S$ must contain $\sigma(\mc S)$ as a subset. $\qed$

This gives a new way of creating $\sigma$-algebras:

• Let $\mc S=\{\{\omega\}\,:\,\omega\in\Omega\}$ be the collection of singleton subsets of $\Omega$. Then any $\sigma$-algebra containing $\mc S$ must contain all countable subsets of $\Omega$, and their complements. Conversely, the collection

  $\Sigma=\{A\subseteq\Omega\,:\,A\text{ or }\Omega\backslash A\text{ countable}\}$

  can be checked to be a $\sigma$-algebra on $\Omega$, so $\sigma(\mc S)=\Sigma$.

• More generally, let $\mc S=\{A_\alpha\,:\,\alpha\in I\}$ be a collection of disjoint subsets of $\Omega$ with union $\Omega$. Then by a similar argument, $\sigma(\mc S)$ consists of all countable unions of $A_\alpha$ and their complements.
• Let $\Omega=\bb R^n$, and let $\mc O_n$ be the collection of open sets in $\bb R^n$. Then $\mc B=\mc B_n=\sigma(\mc O_n)$ contains:
  • all open and closed sets;
  • all $G_\delta$ sets, ie. countable intersections of open sets;
  • all $F_\sigma$ sets, ie. countable unions of closed sets;
  • all $G_{\delta\sigma}$ sets, ie. countable unions of $G_\delta$ sets;
  • all $F_{\sigma\delta}$ sets, ie. countable intersections of $F_\sigma$ sets;
  • …

  The elements of $\mc B$ are called Borel sets.

• Let $a=(a_1,\ldots,a_n)$, $b=(b_1,\ldots,b_n)$, where $-\infty\leq a_k\leq b_k\leq\infty$ for each $k=1,\ldots,n$. Define the half-open interval

  $[a,b)\coloneqq\{(x_1,\ldots,x_n)\in\bb R^n\,:\,x_k\in[a_k,b_k)\text{ for all }k=1,\ldots,n\}.$

  Let $\mc H_n$ be the collection of all half-open intervals.

  Proposition: $\mc H_n$ generates the $\sigma$-algebra of Borel sets, ie. $\sigma(\mc H_n)=\mc B_n$.

  Proof: We give the proof for $n=1$; the case $n>1$ proceeds similarly.

  First, note that

  $\mc H_1\subseteq\mc O_1\implies\sigma(\mc H_1)\subseteq\sigma(\mc O_1)=\mc B_1.$

  Now $\sigma(\mc H_1)$ contains

  $\bigcup_{k=1}^\infty\left[a+\frac1k,b\right)=(a,b).$

  Note that the topology on $\bb R$ is generated by a countable basis of open intervals. Hence every open set in $\bb R$ is a countable union of open intervals, which lies in $\sigma(\mc H_1)$. Thus

  $\mc B_1=\sigma(\mc O_1)\subseteq\sigma(\mc H_1),$

  so $\sigma(\mc H_1)=\mc B_1$. $\qed$