Help needed to understand a theorem from measure theory: Approximation by really simple functions

Solution 1:

In Comment 1, you say ``Note that we do not require that $A \in \Sigma$", but in fact the theorem assumes that $\Sigma$ is generated by $\mathcal{A}$, so $\mathcal{A} \subseteq \Sigma$.

In the first sentence of (P0), we can say a little more: it is actually sufficient just to prove the theorem when $f := \chi_C$ with $C\in\Sigma$ a set of finite measure, i.e. $C\in\tilde\Sigma$.

As you suggest in (P0), it is true that a $\sigma$-algebra does not have to contain infinitely many sets; for example $\Omega$ could be a finite set and $\Sigma$ its power set. If $\Sigma$ only contains finitely many sets, then the assumptions imply that already $\mathcal{A}=\Sigma$, so the theorem just reduces to saying that an integrable function can be approximated in $L^1$ by a simple function, which we already knew.

In (P3), if we knew $\mathcal B=\tilde\Sigma$, then for any $C\in \tilde\Sigma$ and any $\epsilon>0$ we could find a $A_\epsilon\in\mathcal A$ with $\mu(C \bigtriangleup A_\epsilon)<\epsilon$, which is equivalent to the statement $\int_\Omega |f-h_\epsilon|\ d\mu<\epsilon$ with $f:=\chi_C$ and $h\epsilon = \chi_{A_\epsilon}$, which is exactly what we said in (P0) was needed for the proof.

In (P2'), if $\mathcal{B}$ is a monotone class containing $\mathcal A'$, it certainly contains $\mathcal M(\mathcal A')$, the smallest monotone class containing $\mathcal A'$. The monotone class theorem says that $\mathcal M(\mathcal A')=\sigma(\mathcal A')$, so this implies $\mathcal B \supseteq \sigma(\mathcal A')$; this is the direction of containment which is needed to complete the proof. However, in fact, since $\mathcal B$ was (re)defined as a subcollection of $\sigma(\mathcal A')$, the reverse containment also holds, so $B=\sigma(\mathcal A')$.

In (P4'), by the definition of $\mathcal B$, for any set $B \in \sigma(\mathcal A')$ there is some $A\in\mathcal A'$ such that $\mu(B \bigtriangleup A)<\epsilon/2$. You just need to show that $C'\in\sigma(\mathcal A')$, which isn't too hard (Define $\mathcal C$ to be the collection of sets $A\in \mathcal \Sigma$ such that $A\cap \Omega' \in \sigma(\mathcal A')$, and show that $\mathcal C$ is a $\sigma$-algebra and that it contains $\mathcal A$, hence $C \in \mathcal C$.)