$
\def\P{\mathsf{\sf P}}
\def\E{\mathsf{\sf E}}
\def\Var{\mathsf{\sf Var}}
\def\Cov{\mathsf{\sf Cov}}
\def\std{\mathsf{\sf std}}
\def\Cor{\mathsf{\sf Cor}}
\def\R{\mathbb{R}}
\def\c{\,|\,}
\def\bb{\boldsymbol}
\def\diag{\mathsf{\sf diag}}
\def\defeq{\stackrel{\tiny\text{def}}{=}}
$

We attempt in this book to circumvent the use of measure theory as much as possible. However, in several places where measure theory is essential we make an exception (for example the limit theorems in Chapter 8 and Kolmogorov's extension theorem in Chapter 6). Sections containing such exceptions are marked by an asterisk.

This chapter contains a introduction to the parts of measure theory that are the most essential to probability theory. The next chapter covers Lebesgue integration and the Lebesgue measure.

We make the following comments.

- By induction, if $\mathcal{C}$ is an algebra then $A_1,\ldots,A_k\in\mathcal{C}$ implies $A_1\cup\cdots\cup A_n\in\mathcal{C}$.
- Using De-Morgan's rule and the second property above, $A_1,\ldots,A_k\in\mathcal{C}$ implies $A_1\cap\cdots\cap A_n\in\mathcal{C}$,
- A more compact definition of an algebra is a set containing $\Omega$ that is closed under finite unions, finite intersections, and complements.

In other words, if $A_n, n\in\mathbb{N}$ reside in a $\sigma$-algebra $\mathcal{A}$ then we also have $\cup_{n\in\mathbb{N}} A_n\in\mathcal{A}$, $\cap_{n\in\mathbb{N}} A_n\in\mathcal{A}$ and $A_n^c\in\mathcal{A}$.

Note that every $\sigma$-algebra necessarily includes $\emptyset$ and $\Omega$ since $A_n\cap A_n^c=\emptyset$ and $A_n\cup A_n^c=\Omega$. As a consequence, a $\sigma$-algebra is also closed under finite unions and intersections (define $A_k$ above for $k\geq c$ to be either $\emptyset$ or $\Omega$), implying that a $\sigma$ algebra is also an algebra.

In other words, there is a $\sigma$-algebra containing $\mathcal{A}$ that is a subset of all other $\sigma$-algebras containing $\mathcal{A}$.

It is desirable to have an efficient mechanism for producing the smallest $\sigma$-algebra containing a set of sets $\mathcal{C}$. A conceptual mechanism is as follows: (i) create a set $\mathcal{C}'$ containing the sets in $\mathcal{C}$, their complements, and their countable unions and intersections, (ii) repeat step (i) with $\mathcal{C}'$ substituting $\mathcal{C}$. Unfortunately, the mechanism above is not computationally efficient as it may never terminate. The proposition above asserts the existence of a smallest $\sigma$-algebra but does not offer any assistance into constructing it.

We show that the set \[\mathcal{G}=\{A:A\cap U\in\sigma(\mathcal{C}\cap U)\}\] is a $\sigma$-algebra. The first requirement holds since $\Omega\in\mathcal{G}$. If $A\in\mathcal{G}$ then $A\cap U\in\sigma(\mathcal{C}\cap U)$, implying that \[A^c\cap U=U\setminus (A\cap U)\in\sigma(\mathcal{C}\cap U)\] as well (the set difference of two sets in a $\sigma$ algebra is in the $\sigma$ algebra as well). It follows that $A^c\in\mathcal{G}$, showing that the second requirement holds as well. To show the third requirement, note that if $A_n\in\mathcal{G}$ for $n\in\mathbb{N}$ then \[(\cup_n A_n)\cap U = \cup_n (A_n\cap U)\in\sigma(\mathcal{C}\cap U)\] (since $A_n\cap U\in\sigma(\mathcal{C}\cap U)$) and thus $\cup_n A_n\in\mathcal{G}$.

Since $\mathcal{G}$ is a $\sigma$-algebra and $\mathcal{C}\subset \mathcal{G}$ (if $A\in\mathcal{C}$ then $A\cap U\in \sigma(\mathcal{C}\cap U)$ implying that $A\in\mathcal{G}$) we have \[\sigma(\mathcal{C})\subset \sigma(\mathcal{G}) = \mathcal{G}.\] (The last equality holds since $\mathcal{G}$ is a $\sigma$-algebra.) If $A\in\sigma(\mathcal{C})$ then $A\in\mathcal{G}$ and $A\cap U\in\sigma(\mathcal{C}\cap U)$ implying that $A\in \sigma(\mathcal{C}\cap U)$. It follows that $\sigma(\mathcal{C}) \cap U \subset \sigma(\mathcal{C}\cap U)$.