Sigma-Algebra Extended by Single Set

Theorem
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:

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:

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:


 * $\ds \map \sigma {\Sigma \cup \set S} = \Sigma'$