# 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:

 $\displaystyle \left({\left({E_1 \cap S}\right) \cup \left({E_2 \cap S^c}\right)}\right)^c$ $=$ $\displaystyle \left({E_1 \cap S}\right)^c \cap \left({E_2 \cap S^c}\right)^c$ De Morgan's Laws: Difference with Union $\displaystyle$ $=$ $\displaystyle \left({E_1^c \cup S^c}\right) \cap \left({E_2^c \cup S}\right)$ De Morgan's Laws: Difference with Intersection, Set Difference with Set Difference $\displaystyle$ $=$ $\displaystyle \left({ \left({E_1^c \cup S^c}\right) \cap E_2^c }\right) \cup \left({ \left({E_1^c \cup S^c}\right) \cap S }\right)$ Intersection Distributes over Union $\displaystyle$ $=$ $\displaystyle \left({E_1^c \cap E_2^c}\right) \cup \left({E_2^c \cap S^c}\right) \cup \left({E_1^c \cap S}\right) \cup \left({S^c \cap S}\right)$ Union Distributes over Intersection $\displaystyle$ $=$ $\displaystyle \left({ \left({E_1^c \cap E_2^c}\right) \cap \left({S^c \cup S}\right) }\right) \cup \left({E_2^c \cap S^c}\right) \cup \left({E_1^c \cap S}\right)$ Union with Relative Complement, Set Difference Intersection with Second Set is Empty Set $\displaystyle$ $=$ $\displaystyle \left({E_1^c \cap E_2^c \cap S}\right) \cup \left({E_1^c \cap S}\right) \cup \left({E_1^c \cap E_2^c \cap S^c}\right) \cup \left({E_2^c \cap S^c}\right)$ Intersection Distributes over Union $\displaystyle$ $=$ $\displaystyle \left({\left({\left({E_1^c \cap E_2^c}\right) \cup E_1^c}\right) \cap S}\right) \cup \left({\left({\left({E_1^c \cap E_2^c}\right) \cup E_2^c}\right) \cap S^c}\right)$ Intersection Distributes over Union $\displaystyle$ $=$ $\displaystyle \left({E_1^c \cap S}\right) \cup \left({E_2^c \cap S^c}\right)$ Intersection is Subset, Union with Superset is Superset

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:

 $\displaystyle \bigcup_{n \mathop \in \N} \left({E_{1, n} \cap S}\right) \cup \left({E_{2, n} \cap S^c}\right)$ $=$ $\displaystyle \left({\bigcup_{n \mathop \in \N} \left({E_{1, n} \cap S}\right)}\right) \cup \left({\bigcup_{n \mathop \in \N} \left({E_{2, n} \cap S^c}\right)}\right)$ Union Distributes over Union/Families of Sets $\displaystyle$ $=$ $\displaystyle \left({\left({\bigcup_{n \mathop \in \N} E_{1, n} }\right) \cap S}\right) \cup \left({\left({\bigcup_{n \mathop \in \N} E_{2, n} }\right) \cap S^c}\right)$ Union Distributes over Intersection

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

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

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

It follows that:

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

$\blacksquare$