A question in probability theory
If $(\Omega, \mathcal{A}, P)$ is a probability space without atoms, then for every $a \in[0,1]$ there exists at least one set $A \in \mathcal{A}$ of probability $P(A)=a$. The question comes with a hint which I don't understand at all. Hint: Let $\hat{B}_{0}$ be a maximal element of the subclass $\mathcal{B}$ of $\mathcal{A} / P$ consisting of those $\hat{B}$ such that $P(\hat{B}) \leq a$, which subclass is inductive under inclusion. Show that $P(\hat{B})<a$ would imply that the subclass $\mathcal{C}$ of $\mathcal{A} / P$ of those $\hat{C}$ such that $P(\hat{C})>0$ and $\hat{C} \cap \hat{B}=\emptyset$ is inductive for $\supset$; but every maximal element of $\mathcal{C}$ can only be an atom.
Answer
We prove this by contradiction.
Suppose there are no sets $A \in \mathcal{A}$ with $P(A)=a.$ Define
\[\mathcal{B}=\{\hat{B} \in \mathcal{A}: P(\hat{B})\leq a\}.\]
Note that $\mathcal{B}$ is inductive under inclusion, i.e.
\[\hat{B_1}\leq \hat{B_2} \text{iff} \hat{B_1} \subset\hat{B_2}.\]
Thus by Zorn's lemma (every partially ordered set for which every totally ordered subset has an upper bound contains at least one maximal element) $\mathcal{B}$ has a maximal element $\hat{B}_0\in \mathcal{B}$ with $P(\hat{B}_0 )\leq a$. Since we have assumed that there is no set with probability $a$, we must have
\[P(\hat{B}_0 )<a.\]
Next define
\[\mathcal{C}=\{\hat{C}\in \mathcal{A}: \hat{C}\neq \emptyset, P(\hat{C})\geq 0 \text{and} \hat{C}\cap \hat{B}_0=\emptyset\}.\]
Again $\mathcal{C}$ is inductive under inclusion, i.e.
\[\hat{C_1}\leq \hat{C_2} \text{iff} \hat{C_2} \subset \hat{C_1}.\]
Thus by Zorn's lemma it has a maximal element $\hat{C}_0 \in \mathcal{C}$ with
\[P(\hat{C}_0)\geq 0.\]
Now we consider two cases:
Case I: $P(\hat{C}_0)=0$. Since $\hat{C}_0 \neq \emptyset$, and $\hat{C}_0 \cap \hat{B}_0=\emptyset$, we have
\[\hat{B}_0 \subset B^{*}:=\hat{B}_0\cup \hat{C}_0 \]
and
\[P(B^*)=P(\hat{B}_0)+P(\hat{C}_0)=P(\hat{B}_0)\leq a.\]
Thus $B^* \in \mathcal{B}$ and $\hat{B}_0 \subset B^*$. This contradicts the assumption that $\hat{B}_0$ is maximal.
Case II: $P(\hat{C}_0)>0$. Since $\hat{C}_0$ is maximal, $\hat{C}_0$ must be an atom, otherwise we can remove an element from $\hat{C}_0$ and get another set $C^*$ with
\[C^* \geq \hat{C}_0,\]
which contradicts the assumption that $\hat{C}_0$ is maximal.
Both cases I and II lead to a contradiction. Therefore, there should exists a set $A\in \mathcal {A}$ with $P(A)=a.$
Note: To use the Zorn's lemma a set must be partially ordered. In the context above we have
\[\hat{B_1}\subset \hat{B}_2 \text{and} \hat{B_2}\subset \hat{B}_3 \Rightarrow \hat{B_1}\subset \hat{B}_3,\]
which means that inclusion is partially ordered, i.e.
\[\hat{B_1}\leq \hat{B}_2 \text{and} \hat{B_2}\leq \hat{B}_3 \Rightarrow \hat{B_1}\leq \hat{B}_3.\]
This is what it means to say "set to be inductive under inclusion" in this context. You can also see that $\mathcal{C}$ is a partial order. You just need to replace $\subset$ with $\supset$. I suggest you to review Zorn's lemma:
https://en.wikipedia.org/wiki/Zorn%27s_lemma

This took me a while to answer. Please consider offering higher bounties in future.

Leave a comment if you need any clarifications.

Thanks for taking the time! Just a quick question as I'm going over your solution. Could you just clarify a bit more to me what it means for the set to be inductive under inclusion?

I added a note at the end of my solution.

Thanks!
 answered
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