Probability space without atoms

Question

I managed to prove that it is impossible to have a probability space $(\Omega,\mathbb{A},P)$ without atoms. Obviously, this is false. Can anyone tell me where my reasoning falls apart?

Here is my proof: Let $\mathbb{B}:=\{B:P(B)\ge a\}$, where $0<a<1$. We can define a reverse inclusion order on $\mathbb{B}$ so that $B_1\le B_2$ if $B_2 \subset B_1$. Let us consider a chain of sets that belong to $\mathbb{B}$ so that $B_i \le B_j\le \ldots$. The intersection of a countable subsequence of the chain, $\cap_{i=1}^nB_i=\tilde{B}$, is an upper bound of the chain. Moreover, since $P(B_i) \ge a$, it follows that $P(\tilde{B})=\lim P(B_i)\ge a$, so that $\tilde{B} \subset \mathbb{B}$.

Therefore, we can apply Zorn's lemma and find a maximal element in $\mathbb{B}$, say $B_0$. However, since the probability space has no atoms, there exists a subset $A$ of $B_0$ such that $0<P(A)<P(B_0)$. But since $A \le B_0$ it follows that $A=B_0$ so that $P(A)=P(B_0)$.

Answer 1

The problem is that $A$ is not in $\mathbb{B}$, so you do not have that $A\leq B_0$.

Answer 2

There are (at least) two errors.

1. The intersetion of finitly many sets of measure $a$ may not have measure $a$, it could be less. (unless $a=1$)

2. Why is this chain of countable length and you do you get some "finite subchain", which is sufficent to consider

Edit: maybe you have a missconception with the termin "chain" in Zorn's Lemma. A chain is a linearly ordered subset of a partial ordering. This does not mean that in can be indexed by the natural numbers.