Theorem: Sigma-Algebra Extended by Single Set

Let $\Sigma$ be a $\sigma$-algebra on a set $X$.
Let $S \subseteq X$ be a subset of $X$.
For subsets $T \subseteq X$ of $X$, denote $T^\complement$ for the set difference $X \setminus T$.
Then:
:$\map \sigma {\Sigma \cup \set S} = \set {\paren {E_1 \cap S} \cup \paren {E_2 \cap S^\complement}: E_1, E_2 \in \Sigma}$
where $\sigma$ denotes generated $\sigma$-algebra.

Proof:
Define $\Sigma'$ as follows:
:$\Sigma' := \set {\paren {E_1 \cap S} \cup \paren {E_2 \cap S^\complement}: E_1, E_2 \in \Sigma}$
Picking $E_1 = X$ and $E_2 = \O$ (allowed by Sigma-Algebra Contains Empty Set), it follows that $S \in \Sigma'$.
On the other hand, for any $E_1 \in \Sigma$, have by Intersection Distributes over Union and Union with Relative Complement:
:$\paren {E_1 \cap S} \cup \paren {E_1 \cap S^\complement} = E_1 \cap \paren {S \cup S^\complement} = E_1 \cap X = E_1$
Hence $E_1 \in \Sigma'$ for all $E_1$, hence $\Sigma \subseteq \Sigma'$.
Therefore, $\Sigma \cup \set S \subseteq \Sigma'$.
Moreover, from Sigma-Algebra Closed under Union, Sigma-Algebra Closed under Intersection and axiom $(2)$ for a $\sigma$-algebra, it is necessarily the case that:
:$\Sigma' \subseteq \map \sigma {\Sigma \cup \set S}$
It will thence suffice to demonstrate that $\Sigma'$ is a $\sigma$-algebra.
Since $X \in \Sigma$, also $X \in \Sigma'$.
Next, for any $E_1, E_2 \in \Sigma$, observe:
{{begin-eqn}}
{{eqn | l = \paren {\paren {E_1 \cap S} \cup \paren {E_2 \cap S^\complement} }^\complement
      | r = \paren {E_1 \cap S}^\complement \cap \paren {E_2 \cap S^\complement}^\complement
      | c = De Morgan's Laws: Difference with Union
}}
{{eqn | r = \paren {E_1^\complement \cup S^\complement} \cap \paren {E_2^\complement \cup S}
      | c = De Morgan's Laws: Difference with Intersection, Set Difference with Set Difference
}}
{{eqn | r = \paren {\paren {E_1^\complement \cup S^\complement} \cap E_2^\complement} \cup \paren {\paren {E_1^\complement \cup S^\complement} \cap S}
      | c = Intersection Distributes over Union
}}
{{eqn | r = \paren {E_1^\complement \cap E_2^\complement} \cup \paren {E_2^\complement \cap S^\complement} \cup \paren {E_1^\complement \cap S} \cup \paren {S^\complement \cap S}
      | c = Union Distributes over Intersection
}}
{{eqn | r = \paren {\paren {E_1^\complement \cap E_2^\complement} \cap \paren {S^\complement \cup S} } \cup \paren {E_2^\complement \cap S^\complement} \cup \paren {E_1^\complement \cap S}
      | c = Union with Relative Complement, Set Difference Intersection with Second Set is Empty Set
}}
{{eqn | r = \paren {E_1^\complement \cap E_2^\complement \cap S} \cup \paren {E_1^\complement \cap S} \cup \paren {E_1^\complement \cap E_2^\complement \cap S^\complement} \cup \paren {E_2^\complement \cap S^\complement}
      | c = Intersection Distributes over Union
}}
{{eqn | r = \paren {\paren {\paren {E_1^\complement \cap E_2^\complement} \cup E_1^\complement} \cap S} \cup \paren {\paren {\paren {E_1^\complement \cap E_2^\complement} \cup E_2^\complement} \cap S^\complement}
      | c = Intersection Distributes over Union
}}
{{eqn | r = \paren {E_1^\complement \cap S} \cup \paren {E_2^\complement \cap S^\complement}
      | c = Intersection is Subset, Union with Superset is Superset
}}
{{end-eqn}}
As $\Sigma$ is a $\sigma$-algebra, $E_1^\complement, E_2^\complement \in \Sigma$ and so indeed:
:$\paren {\paren {E_1 \cap S} \cup \paren {E_2 \cap S^\complement} }^\complement \in \Sigma'$
Finally, let $\sequence {E_{1, n} }_{n \mathop \in \N}$ and $\sequence {E_{2, n} }_{n \mathop \in \N}$ be sequences in $\Sigma$.
Then:
{{begin-eqn}}
{{eqn | l = \bigcup_{n \mathop \in \N} \paren {E_{1, n} \cap S} \cup \paren {E_{2, n} \cap S^\complement}
      | r = \paren {\bigcup_{n \mathop \in \N} \paren {E_{1, n} \cap S} } \cup \paren {\bigcup_{n \mathop \in \N} \paren {E_{2, n} \cap S^\complement} }
      | c = Set Union is Self-Distributive/Families of Sets
}}
{{eqn | r = \paren {\paren {\bigcup_{n \mathop \in \N} E_{1, n} } \cap S} \cup \paren {\paren {\bigcup_{n \mathop \in \N} E_{2, n} } \cap S^\complement}
      | c = Union Distributes over Intersection
}}
{{end-eqn}}
Since $\ds \bigcup_{n \mathop \in \N} E_{1, n}, \bigcup_{n \mathop \in \N} E_{2, n} \in \Sigma$, it follows that:
:$\ds \bigcup_{n \mathop \in \N} \paren {E_{1, n} \cap S} \cup \paren {E_{2, n} \cap S^\complement} \in \Sigma'$
Hence it is established that $\Sigma'$ is a $\sigma$-algebra.
It follows that:
:$\ds \map \sigma {\Sigma \cup \set S} = \Sigma'$
{{qed}}
Category:Sigma-Algebras
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