Sigma-Algebra Extended by Single Set

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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:

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

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:

\(\ds \bigcup_{n \mathop \in \N} \paren {E_{1, n} \cap S} \cup \paren {E_{2, n} \cap S^\complement}\) \(=\) \(\ds \paren {\bigcup_{n \mathop \in \N} \paren {E_{1, n} \cap S} } \cup \paren {\bigcup_{n \mathop \in \N} \paren {E_{2, n} \cap S^\complement} }\) Set Union is Self-Distributive/Families of Sets
\(\ds \) \(=\) \(\ds \paren {\paren {\bigcup_{n \mathop \in \N} E_{1, n} } \cap S} \cup \paren {\paren {\bigcup_{n \mathop \in \N} E_{2, n} } \cap S^\complement}\) Union Distributes over Intersection

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:

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

$\blacksquare$