Showing Probability Measure is Nonatomic Here is the problem I am working on: Let $(\Omega, \mathcal{F}, P)$ be a probability space, and suppose there are independent events $A_1, A_2,...$ such that if $ \alpha_n = \min\{P(A_n), 1 - P(A_n)\}$ then $\sum \alpha_n = \infty$. Show that $P$ is nonatomic.
Here's what I have so far: Set $B_n$ equal to whichever of $A_n$ or $A_n^c$ has smaller probability. Then $P(B_n = \alpha_n.$ Then
$$
P(B_1 \cap \cdots \cap B_n) = \prod \alpha_j \leq \prod (1 - \alpha_j)
\leq e^{\sum \alpha_n} \rightarrow 0.
$$
I have a hint that says I should consider sets of the form 
$$
\{ \omega: \sum_n I_{A_n}(\omega) 2^{-n} \leq x \}
$$
where $I$ is the indicator function of the set  $A_n$. I think the idea is to then show that for any set $B$ the function $P(B \cap A_x)$ is continuous in $x$, which shows that the measure is non-atomic (by the Intermediate Value Theorem), but I don't see how to do this. Could someone help me show this?
 A: Let us first rewrite $\alpha_n$ in a different form, which we will use later, namely observe that 
$$
\tag{A} \alpha_n = \frac{\mathbb{P}(A_n) + \mathbb{P}(A_n^c) - |\mathbb{P}(A_n) - \mathbb{P}(A_n^c)|}{2} = \frac 12 - \left| \mathbb{P}(A_n) - \frac 12  \right|.
$$
From the definition of non-atomic measure (see Wikipedia), we need to show that there is no event $A\in \mathcal{F}$ with the property that if $B\subset A $ such that $B\in \mathcal{F}$ and $\mathbb{P}(B) < \mathbb{P}(A)$ then  $\mathbb{P}(B) = 0$.
Assume for contradiction that there is $A\in\mathcal{F}$ satisfying the above, clearly we must have $\mathbb{P}(A)>0$.
Let $n\in \mathbb{N}$ be any, then $\mathbb{P}(A) = \mathbb{P}(A \cap A_n) + \mathbb{P}(A \cap A_n^c)$. Hence we either get $\mathbb{P}(A\cap A_n) < \mathbb{P}(A) $ or $\mathbb{P}(A\cap A_n^c) < \mathbb{P}(A) $. But in view of the definition of $A$, we can have only 1 of the strict inequalities, since the left-hand side of the either inequality becomes $0$, while $\mathbb{P}(A) > 0$.
We thus get that $\mathbb{N}$ is being partitioned into two sets $N_1$ and $N_2$, where for each $n \in N_1$ we get $\mathbb{P}(A) = \mathbb{P}(A \cap A_n)$ and for each $n \in N_2$ we have $\mathbb{P}(A) = \mathbb{P}(A \cap A_n^c)$.
If $n \in N_1$, we have $\mathbb{P}(A \setminus A_n) \leq \mathbb{P}(A \setminus (A\cap A_n)) = \mathbb{P}(A) - \mathbb{P}(A \cap A_n)  = 0 $, hence without loss of generality we may assume that $A \subset A_n$.
It follows that $A \subset \bigcap\limits_{n \in N_1} A_n $, from which and the independence of $\{A_n\}$ we get
$$
0 < \mathbb{P}(A) \leq \mathbb{P}( \bigcap\limits_{n \in N_1} A_n ) = \prod\limits_{n \in N_1} \mathbb{P}(A_n) = \prod\limits_{n \in N_1} (1 
- \mathbb{P}(A_n^c) ).
$$
Since the last product converges, we get that 
$$
\sum_{n\in N_1}\mathbb{P}(A_n^c) < \infty \tag{1}
$$
A similar reasoning implies that
$$
\sum_{n\in N_2}\mathbb{P}(A_n) < \infty \tag{2}.
$$
In view of $(1)$ we have $\mathbb{P}(A_n) \to 1$ as $n\to \infty $ and $n\in N_1$, hence
from $(A)$ we obtain $\alpha_n = \frac 12 - ( \mathbb{P}(A_n) - \frac 12 ) = 1 - \mathbb{P}(A_n) = \mathbb{P}(A_n^c)$.
In a similar way, from $(2)$ for $n\in N_2 $ we get $\mathbb{P}(A_n) \to 0$ as $n\to \infty$ and $n \in N_2$, hence $(A)$ becomes $\alpha_n = \frac 12 - (\frac 12 - \mathbb{P}(A_n) ) = \mathbb{P}(A_n)$.
We thus get that $\sum_n \alpha_n $ is a sum of two converging series, namely $(1)$ and $(2)$ (note that we might need to disregard some finite number of terms to get $\mathbb{P}(A_n) > 1/2$ on $N_1$ and $\mathbb{P}(A_n)<1/2$ on $N_2$, but that is irrelevant for convergence of the series). It follows that $\sum_n \alpha_n$ converges, which is a contradiction, proving the non-atomicity of $\mathbb{P}$.
