Exercise 1.7.3 (Boolean algebras that are $\sigma$-algebras)

The only difference between Boolean algebras and $\sigma$-algebras is the countable union property; thus, we only verify this particular characteristic.

• If $B$ is a $\sigma$-algebra, then it is closed under any countable unions, in particular under disjoint ones.

• Suppose that $B$ is closed under countable disjoint unions, and let ${(E_n)}_{n\in N}$ be any sequence of sets in $B$. We argue that:

  (i)

  $\forall <!--\; FUNCTION\; APPLICATION\; -->N\in N:{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^N{E}_n={\bigcup }_{n=1}^N\left(E_n\setminus {\bigcup }_{i=1}^{n-1}{E}_i\right)$.

  (ii)

  $\forall <!--\; FUNCTION\; APPLICATION\; -->n\in N:E_n\setminus {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{n-1}{E}_i\in B$.

  (iii)

  $\forall <!--\; FUNCTION\; APPLICATION\; -->n,m\in N,n>m:\left(E_n\setminus {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{n-1}{E}_i\right)\cap \left(E_m\setminus {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{m-1}{E}_i\right)=\varnothing$.

  If proven, these facts will combined guarantee that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->\{E_n\}$ can be represented as a countable union of disjoint sets in $B$ and must therefore itself be contained in $B$. We start with the first assertion.

  • induction base. Let $N=2$. Then obviously

    $$E_1\cup E_2=E_1\cup (E_2∕E_1).$$

  • induction step. Now suppose inductively that we have proven the assertion for some $N\in N$. We then have

    $$\bigcup _{n=1}^{N+1}{E}_n=\left(\bigcup _{n=1}^N{E}_n\right)\cup E_{N+1}=\left(\bigcup _{n=1}^N{E}_n\right)\cup \left(E_{N+1}\setminus \bigcup _{n=1}^N{E}_n\right)\stackrel{\mathit{IH}}{\text{=}}\bigcup _{n=1}^N\left(E_n\setminus \bigcup _{i=1}^{n-1}{E}_i\right)\cup \left(E_{N+1}\setminus \bigcup _{n=1}^N{E}_n\right)=\bigcup _{n=1}^{N+1}\left(E_n\setminus \bigcup _{i=1}^{n-1}{E}_i\right),$$

    which concludes the induction.

  Hence, we have proven (i). By closedness under finite unions and complements, each $E_n\setminus {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{n-1}{E}_i$ must be contained in the Boolean algebra $B$ and so (ii) is true as well. Finally, these sets are disjoint as

  $$\left(E_n\setminus \bigcup _{i=1}^{n-1}{E}_i\right)\cap \left(E_m\setminus \bigcup _{i=1}^{m-1}{E}_i\right)\subseteq \left(E_n\setminus E_m\right)\cap \left(E_m\right)=\varnothing .$$

  Since the assertion is true for all $N\in N$, it must also be true for the infinite unions:

  $$\bigcup _{n=1}^{\infty }{E}_n=\bigcup _{n=1}^{\infty }\left(E_n\setminus \bigcup _{i=1}^{n-1}{E}_i\right)\in B,$$

  and so we are done.