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: {{begin-eqn}} {{eqn | l = \left({\left({E_1 \cap S}\right) \cup \left({E_2 \cap S^c}\right)}\right)^c       | r = \left({E_1 \cap S}\right)^c \cap \left({E_2 \cap S^c}\right)^c       | c = De Morgan's Laws: Difference with Union }} {{eqn | r = \left({E_1^c \cup S^c}\right) \cap \left({E_2^c \cup S}\right)       | c = De Morgan's Laws: Difference with Intersection, Set Difference with Set Difference }} {{eqn | r = \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)       | c = Intersection Distributes over Union }} {{eqn | r = \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)       | c = Union Distributes over Intersection }} {{eqn | r = \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)       | c = Union with Relative Complement, Set Difference Intersection with Second Set is Empty Set }} {{eqn | r = \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)       | c = Intersection Distributes over Union }} {{eqn | r = \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)       | c = Intersection Distributes over Union }} {{eqn | r = \left({E_1^c \cap S}\right) \cup \left({E_2^c \cap S^c}\right)       | c = Intersection is Subset, Union with Superset is Superset }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \bigcup_{n \mathop \in \N} \left({E_{1, n} \cap S}\right) \cup \left({E_{2, n} \cap S^c}\right)       | r = \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)       | c = Union Distributes over Union/Families of Sets }} {{eqn | r = \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)       | c = Union Distributes over Intersection }} {{end-eqn}} 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: :$\displaystyle \sigma \left({\Sigma \cup \left\{{S}\right\}}\right) = \Sigma'$ {{qed}} Category:Sigma-Algebras quvak41smgejuhrktod8xac8k1ncmym