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 $T^c$ for the set difference $X \setminus T$.

Then:


 * $\sigma \left({\Sigma \cup \left\{{S}\right\}}\right) = \left\{{\left({E_1 \cap S}\right) \cup \left({E_2 \cap S^c}\right): E_1, E_2 \in \Sigma}\right\}$

where $\sigma$ denotes generated $\sigma$-algebra.

Proof
Define $\Sigma'$ as follows:


 * $\Sigma' := \left\{{\left({E_1 \cap S}\right) \cup \left({E_2 \cap S^c}\right): E_1, E_2 \in \Sigma}\right\}$

Picking $E_1 = X$ and $E_2 = \varnothing$ (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:


 * $\left({E_1 \cap S}\right) \cup \left({E_1 \cap S^c}\right) = E_1 \cap \left({S \cup S^c}\right) = E_1 \cap X = E_1$

Hence $E_1 \in \Sigma'$ for all $E_1$, hence $\Sigma \subseteq \Sigma'$.

Therefore, $\Sigma \cup \left\{{S}\right\} \subseteq \Sigma'$.

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:


 * $\Sigma' \subseteq \sigma \left({\Sigma \cup \left\{{S}\right\}}\right)$

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, $E_1^c, E_2^c \in \Sigma$ and so indeed:


 * $\left({\left({E_1 \cap S}\right) \cup \left({E_2 \cap S^c}\right)}\right)^c \in \Sigma'$

Finally, let $\left({E_{1,n}}\right)_{n \in \N}$ and $\left({E_{2,n}}\right)_{n \in \N}$ be sequences in $\Sigma$.

Then:

Since $\displaystyle \bigcup_{n \mathop \in \N} E_{1,n}, \bigcup_{n \mathop \in \N} E_{2,n} \in \Sigma$, it follows that:


 * $\displaystyle \bigcup_{n \mathop \in \N} \left({E_{1,n} \cap S}\right) \cup \left({E_{2,n} \cap S^c}\right) \in \Sigma'$

Thence, it is established that $\Sigma'$ is a $\sigma$-algebra.

It follows that:


 * $\displaystyle \sigma \left({\Sigma \cup \left\{{S}\right\}}\right) = \Sigma'$