Exercise 1.4.28 (Approximation by an algebra)

Consider the family of sets

We try to prove it is a $\sigma$-algebra.

$\varnothing \in C$ We know $\varnothing \in A$ and $\mu (\varnothing \mathrm{\Delta }\varnothing )=0$, so $\varnothing \in C$.

Closure under complements Let $E\in C$. Let $𝜀>0$ be arbitrary and $A\in A$ such that $\mu (E\mathrm{\Delta }A)\le 𝜀$. Then

Since $A^c\in A$, we have $E^c\in C$.

Closure under finite unions By induction, it suffices to show the result for two sets. Let $E_1,E_2\in C$ and $𝜀>0$ be arbitrary. Let $A_1,A_2\in A$ such that

Then

and by monotonicity and the union bound:

Hence $(E_1\cup E_2)\in C$.

Closure under countable unions Now let ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{E}_i$ be a sequence of elements of $C$, and let ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{L}_i$ be their lacunae $(L_i=E_i\setminus {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{j=1}^{i-1}{E}_j)$ , which we can construct because finite unions, finite intersections, and subtractions are possible in $C$. Let $𝜀>0$ be arbitrary. For each $i$, let $A_i\in A$ such that $\mu (L_i\mathrm{\Delta }{A}_i)\le 𝜀2^{-i}$. This construction gives:

Define $M_k={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^k\mu (L_i)$ and consider the sequence ${(M_k)}_{k=1}^{\infty }$. Since ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }\mu (L_i)<\infty$, $(M_k)$ is an increasing sequence converging to $M$. Thus, there exists $m$ such that for all $k\ge m$:

By disjointness and countable additivity, we have:

Let $A^{\prime }={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^m{A}_i\in A$. Let $A^{\prime }={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^m{A}_i\in A$, and we prove the following claim:

Let $x\in \left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{L}_i\right)\mathrm{\Delta }{A}^{\prime }=\left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{L}_i\setminus {\bigcup }_{i=1}^m{A}_i\right)\cup \left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^m{A}_i\setminus {\bigcup }_{i=1}^{\infty }{L}_i\right).$

Assume $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{L}_i\setminus {\bigcup }_{i=1}^m{A}_i$.

• If $x\in L_j$ with $j>m$, then $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=m+1}^{\infty }{L}_i$ and is thus present on the right-hand side.
• If $x\in L_j$ with $j\le m$, then $x\in L_j$ and $x\notin A_j$ because $x\notin {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^m{A}_i$. Hence, $x\in L_j\setminus A_j$, and thus $x$ is covered by ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{j=1}^m(L_j\mathrm{\Delta }{A}_j)$ on the right-hand side.

Assume $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^m{A}_i\setminus {\bigcup }_{i=1}^{\infty }{L}_i$. Then $x\in A_j$ for some $j\le m$, and $x\notin L_j$ for any $j\in \mathbb{N}$, particularly $x\notin L_j$. Hence, $x\in A_j\setminus L_j$, and $x$ is covered by ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{j=1}^m(L_j\mathrm{\Delta }{A}_j)$ on the right-hand side.

Then

and by sub-additivity:

Using the bounds above, this is at most $2𝜀$. Since $𝜀>0$ was arbitrarily small, so is $2𝜀$ and ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{E}_i\in C$.

We have shown that $C$ is a $\sigma$-algebra containing $A$. Thus, it is one of the $\sigma$-algebras intersected over when creating $⟨A⟩$, and hence $⟨A⟩\subseteq C$. The claim follows by applying the property of $C$ on $⟨A⟩$.