Definition. An algebra of sets on \(X\) is a collection \(\mathcal{A}\subseteq\mathcal P(X)\) that satisfies
\(\mathcal{A} \neq \varnothing\).
If \(E \in \mathcal{A}\), then \(E^c \in \mathcal{A}\) (closed under complements).
If \(E_1,E_2,\ldots,E_N \in \mathcal{A}\), then \(\bigcup_{j=1}^{N}E_j \in \mathcal{A}\) (closed under finite unions).
Example. \(\mathcal{A}=\{\varnothing, X, E,E^c\}\) is an algebra.
\(\mathcal{A} \neq \varnothing\).
If \(E \in \mathcal{A}\), then \(E^c \in \mathcal{A}\), (closed under complements).
If \(E_1,E_2,\ldots \in \mathcal{A}\), then \(\bigcup_{j=1}^{\infty}E_j \in \mathcal{A}\), (closed under infinite unions).
The pair \((X, \mathcal A)\) is then called a measurable space.
If \(X\) is any set then \(\{\varnothing, X\}\) and \(\mathcal{P}(X)\) are \(\sigma\)-algebras.
If \(X\) is uncountable, then $$\mathcal{A}=\{E \subseteq X: E \text{ is countable or }E^c \text{ is countable}\}$$ is \(\sigma\)-algebra of countable and co-countable sets.
Algebras (resp. \(\sigma\)-algebras) are also closed under finite (resp. countable) intersections, since \(\big(\bigcup_{j=1}^{\infty}E_j\big)^{c}=\bigcap_{j=1}^{\infty} E_j^{c}\).
If \(\mathcal{A}\) is an algebra, then \(\varnothing \in \mathcal{A}\) and \(X\in \mathcal{A}\). Take \(E \in \mathcal{A}\neq\varnothing\), then \(E^c \in \mathcal{A}\), and consequently \(\varnothing=E \cap E^c \in \mathcal{A}\) and \(X=E \cup E^c \in \mathcal{A}\).
An algebra \(\mathcal{A}\) is a \(\sigma\)-algebra provided it is closed under countable disjoint unions. Indeed, let \(F_1=E_1\) and \[F_k=E_k \setminus \bigg(\bigcup_{j=1}^{k-1}E_{j}\bigg)=E_k \cap \bigg(\bigcup_{j=1}^{k-1}E_j\bigg)^c \quad \text{ for any } \quad k\in\mathbb N.\] Then the \(F_k\)’s belong to \(\mathcal{A}\) and are disjoint, and \[\bigcup_{j=1}^{\infty}E_j=\bigcup_{j=1}^{\infty}F_j.\]
The intersection of any family of \(\sigma\)-algebras on \(X\) is again a \(\sigma\)-algebra.
Definition. If \(\mathcal{E} \subseteq\mathcal{P}(X)\), then \[\sigma(\mathcal{E})=\bigcap_{\substack{\mathcal{A} \supseteq \mathcal{E}\\ \mathcal{A}\text{ is a }\sigma\text{-algebra}}}\mathcal A\] is the smallest \(\sigma\)-algebra containing \(\mathcal{E}\).
\(\sigma(\mathcal{E})\) is called the \(\sigma\)-algebra generated by \(\mathcal{E}\) and is unique.
Example. The family \(\mathcal{E} = \{\{x\}: x \in X\}\) generates \[\sigma(\mathcal E)=\{E \subseteq X: E \text{ is countable or } E^c \text{ is countable}\}.\]
Proof. We know that \[\sigma(\mathcal{E})=\bigcap_{\substack{\mathcal{A} \supseteq \mathcal{E}\\ \mathcal{A}\text{ is a }\sigma\text{-algebra}}}\mathcal A \subseteq \sigma(\mathcal{F})\] $$\tag*{$\blacksquare$}$$
It is also easy to see that \(\sigma(\mathcal E)=\sigma(\sigma(\mathcal E))\). Indeed, \[\sigma(\sigma(\mathcal{E}))=\bigcap_{\substack{\mathcal{A} \supseteq \sigma(\mathcal{E})\\ \mathcal{A}\text{ is a }\sigma\text{-algebra}}}\mathcal A \subseteq \sigma(\mathcal{E})\subseteq \sigma(\sigma(\mathcal{E})).\]
Definition. If \(X\) is any metric space, or more generally any topological space, the \(\sigma\)-algebra generated by the family of open sets in \(X\) (or, equivalently, by the family of closed sets in \(X\)) is called the Borel \(\sigma\)-algebra on \(X\) and is denoted by \[{\rm Bor}(X) \subseteq\mathcal P(X).\] Its members are called Borel sets.
Definition. \(F_{\sigma}\) and \(G_{\delta}\) sets
A countable intersection of open sets is called a \(G_{\delta}\) set.
A countable union of closed sets is called an \(F_{\sigma}\) set.
A countable union of \(G_{\delta}\) set is called a \(G_{\delta \sigma }\) set.
A countable intersection of \(F_{\sigma}\) sets is called an \(F_{\sigma \delta}\) sets.
the open intervals: \(\mathcal{E}_1=\{(a,b): a<b\};\)
the closed intervals: \(\mathcal{E}_2=\{[a,b]: a<b\};\)
the half-open intervals: \[\mathcal{E}_3=\{(a,b]: a<b\} \quad \text{ or } \quad \mathcal{E}_4=\{[a,b): a<b\};\]
the open rays: \[\mathcal{E}_5=\{(a,\infty): a \in \mathbb{R}\} \quad \text{ or } \quad \mathcal{E}_6=\{(-\infty,a): a \in \mathbb{R}\};\]
the closed rays: \[\mathcal{E}_7=\{[a,\infty): a \in \mathbb{R}\} \quad \text{ or } \quad \mathcal{E}_8=\{(-\infty,a]: a \in \mathbb{R}\}.\]
If \(A\) is countable, then \(\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha}\) is the \(\sigma\)-algebra generated by \[\big\{\prod_{\alpha \in A}E_{\alpha}: E_{\alpha}\in \mathcal{A}_{\alpha}\big\}.\]
Proof. Let \[\mathcal{E}=\{\pi_{\alpha}^{-1}[E_{\alpha}]: E_{\alpha} \in \mathcal{A}_{\alpha}, \ \alpha \in A\},\] and \[\mathcal{F}=\big\{\prod_{\alpha \in A}E_{\alpha}: E_{\alpha }\in \mathcal{A}_{\alpha}\big\}.\] By the definition \(\sigma(\mathcal{E})=\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha}\).
We first show that \(\mathcal{E} \subseteq \mathcal{F} \subseteq\sigma(\mathcal{F})\), this will imply \(\sigma(\mathcal{E}) \subseteq \sigma(\mathcal{F})\).
Indeed, if \(\pi_{\alpha}^{-1}[E_{\alpha}] \in \mathcal{E}\) for some \(\alpha\in A\) and \(E_{\alpha}\in\mathcal A_{\alpha}\), then \[\pi_{\alpha}^{-1}[E_{\alpha}]=\prod_{\beta \in A}E_{\beta},\] where \(E_{\beta}=X_{\beta}\) for \(\alpha \neq \beta\). Thus \(\pi_{\alpha}^{-1}[E_{\alpha}] \in \mathcal{F}\) as desired.
We now show that \(\sigma(\mathcal{F}) \subseteq \sigma(\mathcal{E})\). Is suffices to prove that \(\mathcal{F} \subseteq \sigma(\mathcal{E})\). Note that \[\mathcal{F} \ni \prod_{\alpha \in A}E_{\alpha}=\bigcap_{\alpha \in A}\pi_{\alpha}^{-1}[E_{\alpha}] \in \sigma(\mathcal{E}).\] The fact that \(A\) is countable, is essential in the argument. $$\tag*{$\blacksquare$}$$
Suppose that \(\mathcal{A}_{\alpha}=\sigma(\mathcal E_{\alpha})\) for any \(\alpha \in A\). Then \(\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha}\) is generated by \[\mathcal{F}_1=\{\pi_{\alpha}^{-1}[E_{\alpha}]: E_{\alpha} \in \mathcal{E}_{\alpha}, \ \alpha \in A\}.\] If \(A\) is countable and \(X_{\alpha} \in \mathcal{E}_{\alpha}\) for any \(\alpha \in A\), then \(\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha}\) is generated by \[\mathcal{F}_2=\big\{\prod_{\alpha \in A}E_{\alpha}\::\: E_{\alpha} \in \mathcal{E}_{\alpha}\big\}.\]
Proof. We have \[\sigma(\mathcal{F}_1) \subseteq \sigma \left(\{\pi_{\alpha}^{-1}[E_{\alpha}]: E_{\alpha} \in \mathcal{A}_{\alpha}, \ \alpha \in A\}\right) =\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha}.\] We have to prove that \(\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha} \subseteq \sigma(\mathcal{F}_1)\).
For each \(\alpha \in A\), consider the collection \[\mathcal{G}_{\alpha}=\{E \subseteq X_{\alpha}: \pi_{\alpha}^{-1}[E] \in \sigma(\mathcal{F}_1)\}.\]
By the definition of \(\mathcal F_1\) we have \(\mathcal{E}_{\alpha}\subseteq \mathcal{G}_{\alpha}\), since \[\pi_{\alpha}^{-1}[E] \in \sigma(\mathcal{F}_1)\] for all \(E \in \mathcal{E}_{\alpha}\) and \(\alpha \in A\).
One can easily show that \(G_{\alpha}\) is a \(\sigma\)-algebra on \(X_\alpha\).
Hence \(\mathcal{G}_{\alpha}\) is a \(\sigma\)-algebra containing \(\mathcal{E}_{\alpha}\) thus \(\mathcal{A}_{\alpha}=\sigma(\mathcal{E}_{\alpha}) \subseteq \mathcal{G}_{\alpha}\) for all \(\alpha \in A\), so \(\{\pi_{\alpha}^{-1}[E_{\alpha}]: E_{\alpha} \in \mathcal{A}_{\alpha}, \ \alpha \in A\}\subseteq \sigma(\mathcal{F}_1)\) and consequently \[\bigotimes_{\alpha \in A}\mathcal{A}_{\alpha}=\sigma \left(\{\pi_{\alpha}^{-1}[E_{\alpha}]: E_{\alpha} \in \mathcal{A}_{\alpha}, \ \alpha \in A\}\right) \subseteq \sigma(\mathcal{F}_1)\] as claimed. $$\tag*{$\blacksquare$}$$
Let \(X_1,\ldots,X_N\) be metric spaces and let \(X=\prod_{j=1}^{N}X_j\) be equipped with the product metric. Then \[\bigotimes_{j=1}^{N}{\rm Bor}(X_j) \subseteq {\rm Bor}(X).\] If the \(X_j\)’s are separable, then \[\bigotimes_{j=1}^{N}{\rm Bor}(X_j) = {\rm Bor}(X).\]
Definition. A semi-algebra of sets on \(X\) is a collection \(\mathcal{I}\subseteq \mathcal P(X)\) such that
\(\varnothing \in \mathcal{I}\),
if \(E,F \in \mathcal{I}\), then \(E \cap F \in \mathcal{I}\),
if \(E \in \mathcal{I}\), then \(E^c\) is a finite disjoint union of members of \(\mathcal{I}\).
A family containing \(\varnothing\), \(\mathbb R\) and all closed-open intervals \([a, b)\) with \(-\infty\le a<b\le \infty\) is a semi-algebra.
If we consider two measurable spaces \((X, \mathcal A)\) and \((Y, \mathcal B)\) then the set \(\mathcal A\times \mathcal B=\{A\times B: A\in \mathcal A \text{ and } B\in\mathcal B\}\) is a semi-algebra.
If \(\mathcal{I}\) is a semi-algebra, the collection \(\mathcal{A}\) of finite disjoint unions of members of \(\mathcal{I}\) is an algebra.