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

Question

I have spent quite some time trying to understand the proof of the theorem in the title of the post. Figured better to ask for help. I have tried to post independent questions on the forum but they either get no attention or people answer in another context. For example, the reason why I am interested in the theorem is posted here. So I decided to post the full problem. What follows is a bit long, but the problem (the questions) are well-defined. Please bear with me.

I have several question regarding the proof of the theorem (stated below). I highlighted the questions I have by making them bold (the answers to my questions all start with "EDIT", I have inserted these after the discussions in the answer):

Theorem (Approximation by really simple functions, taken from ``Analysis'', Lieb and Loss, Second edition 2001, American Mathematical Society; page 34, the book can be found here):

Let $(\Omega,\Sigma,\mu)$ be a measure space with $\Sigma$ generated by an algebra $\cal A$. Assume that $\Omega$ is sigma-finite in the sense that it can be covered by countably many sets in $\cal A$. Let $f$ be an integrable function and let $\epsilon>0$. Then there is a really simple function $h_\epsilon$ such that $\int_\Omega\vert f-h_\epsilon\vert d\mu<\epsilon$.

Comment 1: A really simple function is a linear combination of really simple characteristic (indicator) functions on $\cal A$. A really simple indicator function $\chi_A$ is defined as $$ \chi_A (x)=\begin{cases} 0 & x\notin A \\ 1 & x\in A \\ \end{cases} $$ where $A\in\cal A$. Note that we do not require that $A\in\Sigma$!

Comment 2: It is also assumed that we are looking at $\cal A$ that contains only sets of finite measure, $\mu(A)<\infty$ for each $A\in\cal A$.

Comment 3: A reminder. $\cal A$ is only an algebra. The only requirements are that it is closed under finite unions and differences: if $A,B\in\cal A$ then $A\cup B\in\cal A$ and $A\sim B,B\sim A\in\cal A$. Note that it is not a sigma-algebra. For example, it is not required that $\emptyset\in\cal A$ or $\emptyset\in\Omega$, nor do we bother with infinite union/interesection operations (apart from the stated requierement that $\Omega$ can be covered by a countable sequence of unions).

The proof strategy is as follows (no need to go into details at this stage).

(P0) Since $f$ can be approximated by a simple function (characteristic function defined on sets in $\Sigma$), it is sufficient to prove the theorem when $f:=\chi_C$ with $C\in\Sigma$. (I suppose this is the standard procedure in measure theory. Use step functions to approximate $f$ from below, and define the integral as the supremum of such steps functions.) However, a sigma-algebra does not have to contain infinitely many sets? (Hope i am correct, i.e. i am not saying something wrong). Provided I am right, then how can one be sure that the sigma algebra used to define integration is interesting enough so that it can be used to cover (approximate in the step function sense) complicated functions, in principle any function? (More detailed question can be found here.)

(P1) Define the family of sets that can be approximated by elements in $\cal A$, we call it ${\cal B}$. Formally: Let $\cal B$ be a family of sets $B\in\Sigma$ such that $\mu(B)<\infty$ and such that for every $\epsilon>0$ there is an $A_\epsilon\in\cal A$ satisfying the condition $$ \mu(B\Delta A_\epsilon)<\epsilon $$ where $X\Delta Y:=(X\sim Y)\cup(Y\sim X)$ denotes the symmetric difference of sets $X$ and $Y$.

(P2) Show that $\cal B$ is a monotone class.

(P3) Use the monotone class theorem to conclude ${\cal B}=\tilde\Sigma$ where $\tilde\Sigma$ denotes the sets in $\Sigma$ with finite $\mu$-measure and, quoting the authors of the book, "we are done". Why do the authors claim that this would be sufficient to prove the theorem? This implies that the fact that $\Sigma$ is a sigma algebra somehow ensures that elements from it can approximatively cover any subset in $\Omega$????? (More detailed question can be found here.) EDIT: It is the special context of $\mathcal B$ that makes this claim possible: The claim holds by the definition of $\mathcal B$.

However, there is a problem with the above proof strategy since $\Omega\notin\cal A$ so the monotone class theorem cannot be used directly. A bit more work is needed so the structure of the proof is a bit more complicated, but follows the steps (P0-P3) conceptually. So the real proof follows:

(P0') Find a finite union of sets in $\cal A$, we call it $\Omega'$, such that $\mu(C\Delta C')<\epsilon/2$ where $C':=\Omega'\cap C$. Note that $\Omega'$ covers $C$ but not in an approximative way. It can cover much more than what is needed. Such a finite union always exists since elements in $\cal A$ cover $\Omega$ in a countable way. The idea is to keep covering $C$ until we either cover it all, or what is left uncovered (within $C$, not outside of it) is very small. Consider a sub-algebra of $\cal A$ restricted on $\Omega'$, we call it ${\cal A}'$ (it is possible to show that $\cal A'$ is indeed an algebra). Note that now we have by construction $\Omega'\in\cal A'$, which prepares the stage for the use of the monotone convergence theorem.

(P1') Redefine $\cal B$ with $\cal A'$ in place of $\cal A$.

(P2') Show that $\cal B$ is a monotone class. Not hard to do, I understand the procedure in the book. The authors never bother to show that $\cal B$ is the smallest monotone class. Why can they use the monotone class theorem nevertheless? (More detailed question on this can be found here.) EDIT: I was right to be cautious about that. See the answers below.

(P3') Use the monotone class theorem to conclude that $\cal B$ is the smallest sigma-algebra that contains $\cal A'$.

(P4') Based on the fact that $\cal B$ is a sigma algebra conclude that we can find an $A\in\cal A'$ such that $\mu(C'\Delta A)<\epsilon$/2. I do not understand the claim. Why the fact that $\cal B$ is a sigma-algebra ensures that one can approximate any $C'$ with elements from $\cal B$? (More detailed version of the question can be found here.) EDIT: Here I was wrong. It is the special context of $\mathcal B$ that is crucial: The claim is true by definition, but not otherwise (in principle this is what we are trying to prove).

Answer 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 $C \in \mathcal C$.)