Monotone Class Theorem for Functions

Question

Suppose $\mathcal F$ is a collection of real-valued functions on $X$ such that the constant functions are in $\mathcal F$ and $f + g$, $fg$, and $cf$ are in $\mathcal F$ whenever $f, g \in \mathcal F$ and $c \in \mathbb R$. Suppose $f \in \mathcal F$ whenever $f_n \to f$ and each $f_n \in \mathcal F$. Define the function $$\textbf1_A(x) = \begin{cases} 1, & x \in A;\\ 0, & x \notin A. \end{cases}$$ Prove that $\mathcal A = \{A \subseteq X : \textbf 1_A \in \mathcal F\}$ is a $\sigma$-algebra.

I have the following result at my disposal:

Monotone Class Theorem: Suppose $\mathcal A_0$ is an algebra, $\mathcal A$ is the smallest $\sigma$-algebra containing $\mathcal A_0$, and $\mathcal M$ is the smallest monotone class containing $\mathcal A_0$. Then $\mathcal M = \mathcal A$.

Notes: here are some observations I've made:

1. $\emptyset \in \mathcal A$ since $\textbf 1_\emptyset = 0$ is constant and $\mathcal F$ contains all constant functions, similarly $X \in \mathcal A$.
2. If $A \in \mathcal A$, then $\textbf 1_A \in \mathcal F$; notice that $1 - \textbf 1_A = \textbf 1_{A^c} \in \mathcal F$ since finite sums are of elements in $\mathcal F$ are in $\mathcal F$, hence $A^c \in \mathcal A$.
3. Let $A_1, A_2 \in \mathcal A$, then $\textbf 1_{A_1}, \textbf 1_{A_2} \in \mathcal F$. Observe that $\textbf 1_{A_1 \cap A_2} = \textbf 1_{A_1} \cdot \textbf 1_{A_2} \in \mathcal F$. By mathematical induction if $A_1, \ldots, A_n \in \mathcal A$, then $\bigcap_{i = 1}^n A_i \in \mathcal A$.
4. Since we have finite intersections and complements, we get finite unions: $\bigcup_{i = 1}^n A_i \in \mathcal A$.

So $\mathcal A$ is an algebra. At this point we can choose to make $\mathcal A_0 := \mathcal A$ our algebra.

The problem with showing the result directly comes down to showing if $A_i \in \mathcal A$, then $\bigcup_{i = 1}^\infty A_i \in \mathcal A$.

I'm guessing I want to show that $\mathcal M := \mathcal A_0$ is a monotone class. Once we've done this, since $\mathcal A_0 = \mathcal M$, $\mathcal M$ is obviously the smallest monotone class containing $\mathcal A_0$. By the monotone class theorem $\mathcal M = \sigma(\mathcal A_0)$ is a $\sigma$-algebra, correct? (Question 1)

So let's try to show that $\mathcal M$ is a monotone class.

1. Let $A_k \in \mathcal M$ with $A_k \nearrow A$. In particular, we have that $A_1 \subseteq A_2 \subseteq \cdots$ and $A := \bigcup_{k = 1}^\infty A_k$. Observe that (by hypothesis), $$\lim_{n \to \infty} \textbf 1_{A_n} = \textbf 1_{\bigcup_{k = 1}^\infty A_k}$$ and hence $\bigcup_{k = 1}^\infty A_k \in \mathcal A$. Is this step valid? It seems obvious, but I'm not sure if I've made an assumption here that I shouldn't have (Question 2)
2. Let $A_k \in \mathcal M$ with $A_k \searrow A$. In particular, we have that $A_1 \supseteq A_2 \supseteq \cdots$ and $A := \bigcap_{k = 1}^\infty A_k$. Observe that (by hypothesis) $$\lim_{n \to \infty} \textbf 1_{A_n} = \textbf 1_{\bigcap_{k = 1}^\infty A_k}$$ and hence $\bigcap_{k = 1}^\infty A_k \in \mathcal A$.

If the reasoning in the above two steps is valid about the limit being correct, then can I skip having to use the monotone class theorem and just observe the following? (Question 3)

Let $A_k \in \mathcal A$, then $\textbf 1_{A_k} \in \mathcal F$. Define $B_n = \bigcup_{k = 1}^n A_k \in \mathcal A$ since we've already established that it is an algebra and algebras have finite unions. Define $f_n = \textbf 1_{B_n}$ and notice that $\lim_{n \to \infty} f_n = \lim_{n \to \infty} \textbf 1_{B_n} = \textbf 1_{\cup_{k = 1}^\infty A_k} \in \mathcal F$ by hypothesis. Hence $\bigcup_{k = 1}^\infty A_k \in \mathcal A$. Conclude by definition that $\mathcal A$ is a $\sigma$-algebra.

If my reasoning fails somewhere, I would appreciate any hints/full proofs that establish the result using the monotone class theorem, if just because I feel I lack understanding of how to use it and I think this is a good example where it's usable.

Answer 1

Everything you did seems fine to me. For completeness's sake, here's a succinct solution:

Firstly, $\boldsymbol{1}_\varnothing = 0 \in \mathcal F \implies \varnothing \in \mathcal A$. Next for $A \in \mathcal A$, we have $\boldsymbol{1}_{A^c} = 1 - \boldsymbol{1}_{A} \in \mathcal F \implies A^c \in \mathcal A$. And finally, suppose $(A_k)_{k=1}^\infty$ is a sequence of sets in $\mathcal A$. Let us consider the sequence $\left(\prod_{k=1}^n \boldsymbol{1}_{A_k}\right)_{n=1}^\infty$ of functions in $\mathcal F$. If $x \in \bigcap_{k=1}^\infty A_k$, then $\prod_{k=1}^n\boldsymbol{1}_{A_k}(x) = 1$ for all $n$. On the other hand if $x \notin \bigcap_{k=1}^\infty A_k$, then $\prod_{k=1}^n \boldsymbol{1}_{A_k}(x) = 0$ for all sufficiently large $n$. Thus $\lim_{n \to \infty}\prod_{k=1}^n \boldsymbol{1}_{A_k} = \boldsymbol{1}_{\bigcap_{k=1}^\infty A_k}$ pointwise, implying that $\mathcal A$ is closed under countable intersections. It follows that $\mathcal A$ is indeed a $\sigma$-algebra.