Proving a least upper property for $\sigma$-algebras on probability spaces

Question

Let $(\Omega,\mathcal F,P)$ be probability space. Consider the partial order on $\mathcal F$ defined by $A\leq B \iff P[A\setminus B]=0$, i.e. almost sure inclusion. It has the following properties:

1. $A\leq A$ for all $A\in\mathcal F$.
2. If $A\leq B$ and $B\leq A$ then $P[A \triangle B]=0$.
3. If $A\leq B$ and $B\leq C$ then $A\leq C$.

Am reading the following proposition

Proposition. Every non empty collection $\mathcal D ⊂ \mathcal F$ has a supremum. Further, if $\mathcal D$ is directed upwards ($A \cup B \in \mathcal D$ for all $A, B \in \mathcal D$) there is an increasing sequence $(A_n)$ in $\mathcal D$ such that $\cup_{n=1}^\infty A_n$ achieves the supremum almost surely.

Proof. Let $$c=\sup \Big\{P[\cup\mathcal C]: \mathcal C\subset \mathcal D \text{ and } \mathcal C\text{ is countable}\Big\} $$

Since $\sigma$-algebras are closed under countable unions this is well-defined. For each $n$ we can choose countable $\mathcal C_n\subset \mathcal D $ such that

$$c-\frac{1}{n}< P[\cup\mathcal C_n]\leq c$$

Let $B=\cup [\cup_{n=1}^{\infty}\mathcal C_n]$. Since a countable union of countable sets is countable, we have that $B\in \mathcal F$ and that $P[B]\leq c$. Moreover $\cup\mathcal C_n \subset B$ for all $n$, so by monotonicity of measures we get

$$c-\frac{1}{n}< P[\cup\mathcal C_n]\leq P[B]\leq c$$

for all $n$. It follows that $P[B]=c$. I claim that $A$ is an upper bound for $\mathcal D$. Otherwise there would exists $A\in\mathcal D$ such that $P[A\setminus B]>0$. But then $\mathcal C:=\cup_{n=1}^{\infty}\mathcal C_n \cup \{A\}$ would be a countable subset of $\mathcal D$ satisfying

$$P[\cup \mathcal C]=P[B\cup A]>P[B]=c$$

a contradiction. If $B'$ is another upper bound for $\mathcal D$, then $P[C\setminus B']=0$ for all $C\in \cup_{n=1}^{\infty}\mathcal C_n$. Since there are countably many such $C$, and a countable union of null sets is null, we have that $P[B\setminus B']=0$. So $B$ is a least upper bound for $\mathcal D$.

By construction, there is an increasing sequence approximating $B$ if $\mathcal D$ is directed upwards.

Questions:

1. Why do we need $\mathcal D$ nonempty? It seems that the proof shows that the supremum is $\emptyset$ if $\mathcal D$ is empty.

2. Why is the last sentence true? If we define $A_n=\cup [\cup_{k=1}^{n}\mathcal C_k]$ for each $n$, then $B=\cup_{n=1}^{\infty}A_n$ is a countable union of increasing sets in $\mathcal F$. But don't we need $\mathcal D$ to be closed under countable unions if we want $A_n\in\mathcal D$?

Answer 1

1. Note that if $(S,\le)$ is a poset, $\sup \varnothing$ exists if and only if $S$ has the smallest element.
2. You may construct an increasing sequence $\{A_n\}$ in $\mathcal{D}$ as follows. Take $A_1=\mathcal{C}_{1,1}$, $A_2=A_1\cup \mathcal{C}_{2,1}$, $A_3=A_2\cup \mathcal{C}_{1,2}$, $A_4=A_3\cup \mathcal{C}_{3,1}$, $A_5=A_4\cup \mathcal{C}_{2,2}$, and so on.