Sigma-algebra

Proposition 1

\[\begin{eqnarray} \mathcal{A} := \{ A_{1} \times \cdots \times A_{n} \mid A_{i} \in \mathcal{B}(\mathbb{R}) \} . \end{eqnarray}\]

(1)

\[\begin{eqnarray} A & := & A_{1} \times \cdots \times A_{n} \in \mathcal{A}, \nonumber \\ B & := & B_{1} \times \cdots \times B_{n} \in \mathcal{A}, \nonumber \\ A \cap B & = & (A_{1} \cap B_{1}) \times \cdots \times (A_{n} \cap B_{n}) \nonumber \\ A^{c} & = & A_{1}^{c} \times \cdots \times A_{n}^{c} \nonumber \end{eqnarray}\]

(2)

\[\begin{eqnarray} A & := & A_{1} \times \cdots \times A_{n} \in \mathcal{A}, \nonumber \\ B & := & B_{1} \times \cdots \times B_{n} \in \mathcal{A}, \nonumber \\ A \cap B & \in & \mathcal{A} \nonumber \\ A^{c} & \in & \mathcal{A} \nonumber \\ A \cup B & \in & \mathcal{A} \nonumber \end{eqnarray}\]

proof

proof of (1)

\[\begin{eqnarray} & & (a_{1}, \ldots, a_{n}) \in (A \cap B) \nonumber \\ & \Leftrightarrow & a_{i} \in A_{i}, a_{i} \in B_{i}, \nonumber \\ & \Leftrightarrow & a_{i} \in (A_{i} \cap B_{i}), \nonumber \\ & \Leftrightarrow & (a_{1}, \ldots, a_{n}) \in (A_{1} \cap B_{1}) \times \cdots \times (A_{n} \cap B_{n}), \nonumber \end{eqnarray}\] \[\begin{eqnarray} & & (a_{1}, \ldots, a_{n}) \in A^{c} \nonumber \\ & \Leftrightarrow & a_{i} \notin A_{i} \nonumber \\ & \Leftrightarrow & a_{i} \in A_{i}^{c} \nonumber \\ & \Leftrightarrow & (a_{1}, \ldots, a_{n}) \in A_{1}^{c} \times \cdots \times A_{n}^{c} \nonumber \end{eqnarray}\]

proof of (2)

These are Immediate consequence from (1).

$\Box$

Remark

Note that

\[\begin{eqnarray} (a_{1}, \ldots, a_{n}) \in A \cup B & = & \left( \bigcap_{i \in [1:n]} (a_{i} \in A_{i}) \right) \cup \left( \bigcap_{i \in [1:n]} (b_{i} \in B_{i}) \right) \nonumber \\ & = & \bigcap_{i \in [1:n]} \left( (a_{i} \in A_{i}) \cup \left( \bigcap_{j \in [1:n]} (b_{j} \in B_{j}) \right) \right) \nonumber \\ \nonumber \\ & = & \bigcap_{i \in [1:n]} \bigcap_{j \in [1:n]} \left( (a_{i} \in A_{i}) \cup (b_{j} \in B_{j}) \right) \nonumber \end{eqnarray}\]

■

Proposition 2

• $S$,
  • set
• $\mathcal{A}$,
  • set of subsets of $S$
• $A^{c}$,
  • complement of $A$,
  • $A \subseteq S$,
• $A^{cc} = A$,
  • $A \subseteq S$,

\[\begin{eqnarray} \mathfrak{M} & := & \left\{ \bigcap_{i \in \mathbb{N}} A_{i}^{k_{i}} \mid A_{i} \in (\mathcal{A} \cup \{S\}), \ k_{i} \in \{c, cc\} \right\} . \nonumber \\ \mathcal{F} & := & \left\{ \bigcup_{i \in \mathbb{N}} B_{i}^{k_{i}} \mid B_{i} \in \mathfrak{M}, \ k_{i} \in \{c, cc\} \right\} \nonumber \end{eqnarray}\]

Then

\[\sigma(\mathcal{A}) = \mathcal{F} .\]

TODO: reconsider statement.

proof

\[\begin{equation} B_{i} \in \mathfrak{M} \Rightarrow B_{i}, (B_{i})^{c} \in \mathcal{F} \label{proposition_02_equation_01} . \end{equation}\]

($\subseteq$)

By definition, $\mathcal{A} \subseteq \mathcal{F}$. If $\mathcal{F}$ is $\sigma$-algebra, the inclusion holds.

(1) $S, \emptyset \in \mathcal{F}$.

Indeed, by definiton, $S \in \mathcal{F}$. Hence

\[\bigcup_{i \in \mathbb{N}} S^{c} \bigcap_{i \in \mathbb{N}} S^{c} = \emptyset \in \mathcal{F} .\]

(2) $A \in \mathcal{F} \Rightarrow A^{c} \in \mathcal{F}$.

Indeed, let

\[\begin{eqnarray} \bigcup_{i \in \mathbb{N}} \left( B_{i} \right)^{l_{i}} & \in & \mathcal{F} \nonumber \\ \bigcap_{j \in \mathbb{N}} A_{i,j}^{k_{i,j}} & =: & B_{i} \end{eqnarray}\]

be fixed.

\[\begin{eqnarray} \left( \bigcup_{i \in \mathbb{N}} \left( \bigcap_{j \in \mathbb{N}} A_{j}^{k_{j}} \right)^{l_{i}} \right)^{c} & = & \bigcap_{i \in \mathbb{N}} \left( \bigcap_{j \in \mathbb{N}} A_{j}^{k_{j}} \right)^{l_{i}c} . \nonumber \end{eqnarray}\]

By \(\eqref{proposition_02_equation_01}\), we obtain

\[\left( \bigcap_{j \in \mathbb{N}} A_{j}^{k_{j}} \right)^{l_{i}c} \in \mathcal{F} .\]

($\supseteq$)

Let $\bigcap_{i \in \mathbb{N}} A_{i}^{k_{i}} \in \mathfrak{M}$ be fixed. Since \(\sigma(\mathcal{A}) \supseteq (\mathcal{A} \cup \{S\})\),

\[A_{i}^{k_{i}} \in \sigma(\mathcal{A}) .\]

Since $\sigma$-algebra is closed under countable infinite intersection,

\[\bigcap_{i \in \mathbb{N}} A_{i}^{k_{i}} \in \sigma(\mathcal{A}) .\]

By definition of $\sigma$-algebra, it is closed under countable infinite uninon. Thus,

\[\mathcal{F} \subseteq \sigma(\mathcal{A}) .\]

$\Box$

Reference