There exist on some discrete probability space independent events $A_n$ satisfying $P\left(A_n\right)=p_n$. 
Suppose that $0\leq p_n\leq 1$ and put
$\alpha_n=\min\left\{p_n,\,1-p_n\right\}$.
Show that, if $\sum\alpha_n$ converges, then on some
discrete probability space there exist independent events $A_n$ satisfying
$P\left(A_n\right)=p_n$.

This is problem 5.15 in Billingsley's Probability and Measure.
In this section, it has already proved that for any sequence of measures $\left\{\mu_n\right\}$ on the class of all subsets of $\mathbb{R}$ with finite support, there exist on some probability space a sequence of independent simple random variables $\left\{X_n\right\}$ whose distribution is $\mu_n$.
But the probability space constructed in that proof is just the unit interval and I am unable to modify it to meet the requirement of this problem.
 A: Originally, the question was asked without the "independent" qualifier that was present in the Billingsley book. I have updated my answer to account for this requirement.

For simplicity, I will assume without loss of generality that each $p_n\le \frac12$, so that $\alpha_n=p_n$ and $\sum_{n\ge 1}p_n<\infty$.
We proceed the same way that Billingsley did when proving Theorem $5.3$. We start by defining a probability space on the unit interval, which has random variables $X_1,X_2,X_3,\dots$, each taking values in $\{0,1\}$, such that $P(X_n=1)=p_n$. This is done by dividing the unit interval $[0,1]$ in the ratio of $p_1:(1-p_1)$, then dividing both of those pieces in the ratio $p_2:(1-p_2)$, then dividing all four resulting pieces in the ratio $p_3:(1-p_3)$, and so on to infinity.
After all infinitely many stages, there are countably many division points, and removing these division points divides the unit interval into uncountably many connected components. This is because a connected component is specified by an infinite binary sequence, where each entry specifies whether you descend to the left sub-interval or the right sub-interval at each step. The key idea of the proof is that there will be only countably many components which have a positive length, and the total length of all the positive-length components equals $1$, the length of the whole interval. Therefore, you get an equivalent discrete probability space by having one point for each positive-length interval, with probability equal to its length.
I will use the symbols $L$ and $R$ for left and right. A connected component is specified by an infinite sequence $D=(D_1,D_2,D_3,\dots)$, where each $D_i\in \{L,R\}$. For each $n\in \{1,2,\dots\}$, define a function $f_n$ with domain $\{L,R\}$, such that
$$
f_n(L)=p_n,\qquad f_n(R)=1-p_n
$$
It follows that the length of the interval specified by $(D_1,D_2,D_3,\dots)$ is
$$
\prod_{n=1}^\infty f_n(D_n)
$$
This because when you descend left at the $n^\text{th}$ stage, the sub-interval is smaller by a factor of $p_n$, and when you descend right, it shrinks by $1-p_n$. You can then prove the following facts:

*

*If the $D$ sequence contains infinitely many $L$'s, then the length of the resulting interval is zero, because the infinite product $\prod_{n=1}^\infty f_n(D_n)$ will contain infinitely many factors which are at most $1/2$.


*If the $D$ sequence contains only finitely many $L$'s, then the length is positive. For example, the component $(R,R,R,\dots)$ where you descend right each time has length
$$
\prod_{n=1}^\infty(1-p_n)
$$
which is strictly positive because $\sum_{n} p_n<\infty$. In general, changing a finite number of those factors to their complements still leaves a strictly positive product.
All that remains to show is that the total length of all the positive-length components is equal to $1$. To this end, let $q_n=1-p_n$. We need to show that the following infinite sum equals $1$:
$$
\begin{align}
&\hspace{13pt}q_1\cdot q_2\cdot q_3\cdot q_4 \cdots q_n\cdots 
\\&+\color{blue}{p_1}\cdot q_2\cdot q_3\cdot q_4 \cdots 
\\&+q_1\cdot\color{blue}{p_2}\cdot q_3\cdot q_4 \cdots 
\\&+\color{blue}{p_1}\cdot \color{blue}{p_2}\cdot q_3\cdot q_4 \cdots 
\\&+q_1\cdot q_2\cdot \color{blue}{p_3}\cdot q_4 \cdots 
\\&+\color{blue}{p_1}\cdot q_2\cdot \color{blue}{p_3}\cdot q_4 \cdots 
\\&+q_1\cdot \color{blue}{p_2}\cdot \color{blue}{p_3}\cdot q_4 \cdots 
\\&+\color{blue}{p_1}\cdot \color{blue}{p_2}\cdot \color{blue}{p_3}\cdot q_4 \cdots 
\\&\hspace{2cm}\vdots
\end{align}
$$
Fortunately, this not too hard to show. For any natural number $N$, the partial sum of the first $2^N$ terms is exactly equal to
$$
\text{sum of first $2^N$ terms}=\prod_{n=N+1}^\infty q_n \tag{$*$}
$$
Letting $N\to\infty$, this shows the infinite sum equals the limit of the tail-products. Since the infinite product of the $q_n$'s is nonzero, the limit of the tail products is $1$, completing the proof.
Here is the proof of $(*)$ in the simple case $N=1$:
$$
q_1\cdot q_2\cdot q_3\cdots + p_1\cdot q_2\cdot q_3\cdots =(p_1+q_1)\cdot q_2\cdot q_3\cdots = 1\cdot q_2\cdot q_3\cdots 
$$
The same reasoning leads to a proof of $(*)$ for all $N$, using induction (and perhaps some messy bookkeeping if you really want to be rigorous).
A: For every $n$, define the distribution $\mu_n$ by
\begin{align}
\mu_n\{0\} = \alpha_n, \; \mu_n\{1\} = 1 - \alpha_n. 
\end{align}
By Theorem $5.3$, there exists on the probability space $(\Omega, \mathscr{F}, P) := ((0, 1], \mathscr{B}, \lambda)$ an independent sequence $\{X_n\}$ of binary random variables such that $X_n$ has distribution $\mu_n$ (in the construction therein, $P[X_n = 0] = \mu_n\{0\} = \alpha_n$ and $P[X_n = 1] = \mu_n\{1\} = 1 - \alpha_n$ for all $n$).
For a given binary sequence $s_u := (u_1, u_2, \ldots)$, where $u_i \in \{0, 1\}$, the set
\begin{align}
A_{s_u} := [\omega \in \Omega: X_i(\omega) = u_i, i \geq 1] 
= \bigcap_{n \geq 1}[\omega \in \Omega: X_n(\omega) = u_n] \in 
\mathscr{F}.
\end{align}
To evaluate the probability of $A_{s_u}$, we have the following two cases:

*

*There are infinitely many $i$ such that $u_i = 0$, then
\begin{align}
P(A_{s_u}) \leq \prod_{i: u_i = 0}\mu_i\{0\} 
= \prod_{i: u_i = 0}\alpha_i \leq \prod_{i: u_i = 0}
\frac{1}{2} = 0, 
\end{align}
i.e., $P(A_{s_u}) = 0$.

*There exists $N \in \mathbb{N}$, such that $u_i = 1$ for all $i > N$. Define
\begin{align}
A_{s_u}^{(n)} = [\omega \in \Omega: X_i(\omega) = u_i, 1 \leq 
i \leq n].
\end{align}
Then $A_{s_u}^{(n)} \in \mathscr{F}$ and $A_{s_u}^{(n)} \downarrow A_{s_u}$, hence $P(A_{s_u}) = \lim\limits_{n \to \infty}P(A_{s_u}^{(n)})$
by Theorem 2.1(ii).  For $n > N$, by the independence of $X_1, \ldots, X_n$, it follows that
\begin{align}
P(A_{s_u}^{(n)}) = \prod_{i = 1}^nP[X_i = u_i] = \prod_{i = 1}^N P[X_i = u_i]\prod_{i = N + 1}^n (1 - \alpha_i).  
\end{align}
Since $\sum \alpha_n$ converges, the infinite product $\prod\limits_{i = N + 1}^\infty(1 - \alpha_i)$ is strictly positive (see Note 1 below), hence
\begin{align}
P(A_{s_u}) = \lim_{n \to \infty}P(A_{s_u}^{(n)}) = \prod_{i = 1}^N P[X_i = u_i]\prod_{i = N + 1}^\infty (1 - \alpha_i) > 0.  \tag{1}
\end{align}
Now define a new sample space $\Omega'$ of binary sequences by:
\begin{align}
\Omega' = \{\omega' = (u_1, u_2, \ldots): P(A_{s_u}) > 0\}.
\end{align}
Based on the case analysis above, $\Omega'$ is countable. We then let
$\mathscr{F}'$ be the $\sigma$-field of all subsets of $\Omega'$, and define a nonnegative function $p(\omega')$ on $\Omega'$ by $p(\omega') = P(A_{\omega'})$. Define the set function $P'$ on $\mathscr{F}'$ by
\begin{align}
P'(A) = \sum_{\omega' \in A}p(\omega'), \quad A \in \mathscr{F}'.   
\end{align}
It can be verified that $P'$ is indeed a probability measure:

*

*$P'(A) \geq 0$ for $A \in \mathscr{F}'$.  Trivial.


*$P'(\Omega') = 1$.  To prove it, note that
\begin{align}
 P'(\Omega') = \sum_{\omega' \in \Omega'}P(A_{\omega'}) 
= \sum P[X_1 = u_1, X_2 = u_2, \ldots] =: P(S) 
\end{align}
where the last summation extends over all binary sequences $(u_1, u_2, \ldots)$ such that $P[X_1 = u_1, X_2 = u_2, \ldots] > 0$ (so that the event $S$ is the disjoint union of sets $A_{(u_1, u_2, \ldots)}$ with $P(A_{(u_1, u_2, \ldots)}) > 0$). Note that the complementary set of $S$ is $[X_n = 0 
\text{ i.o.}]$. Since $\sum_n P[X_n = 0] = \sum \alpha_n < \infty$, the first Borel-Cantelli lemma gives $P(S^c) = P[X_n \text{ i.o.}] = 0$, hence
$P'(\Omega') = P(S) = 1 - P(S^c) = 1$.


*$P'$ is countably additive.  Trivial.
Therefore, $(\Omega', \mathscr{F}', P')$ is a discrete probability space. Now define events $A_n \subset \Omega'$ as follows: for $n = 1, 2, \ldots$,
\begin{align}
A_n = \begin{cases}
[(u_1, \ldots, u_{n - 1}, 0, u_{n + 1}, \ldots)] & \text{if } \alpha_n = p_n, \\[1em]
[(u_1, \ldots, u_{n - 1}, 1, u_{n + 1}, \ldots)] & \text{if } \alpha_n = 1 - p_n.  
\end{cases} \tag{2}
\end{align}
For fixed $n$, let $u_n^* \in \{0, 1\}$ be the common $n$-th term of sequences in $A_n$, then (the summation extends over $(u_1, \ldots, u_{n - 1}, u_{n + 1}, \ldots)$ such that $P[X_n = u_n^*, X_i = u_i, i \neq n] > 0$):
\begin{align}
 & P'(A_n) \\
=& \sum P[X_1 = u_1, \ldots, X_{n - 1} = u_{n - 1}, X_n = u_n^*, X_{n + 1} = u_{n + 1}, \ldots] \\
\overset{*1}{=}& \sum p_nP[X_1 = u_1, \ldots, X_{n - 1} = u_{n - 1}, X_{n + 1} = u_{n + 1}, \ldots] \\
\overset{*2}{=}& p_n.  
\end{align}
In the above derivation, the equality "$\overset{*1}{=}$" holds because $\sigma(X_n)$ is independent of $\sigma(X_i, i \neq n)$ in $(\Omega, \mathscr{F}, P)$. The equality "$\overset{*2}{=}$" can be derived as in arguing $P'(\Omega') = 1$ above by the first Borel-Cantelli lemma.  See Note 2 below for details.
Lastly, $A_1, A_2, \ldots$ defined in $(2)$ are independent: for any fixed $n \in \mathbb{N}$, let $u_i^*$ be the $i$-th term of the sequences in $A_i, 1 \leq i \leq n$, then $A_1 \cap A_2 \cap \cdots \cap A_n = [(u_1^*, \ldots, u_n^*, u_{n + 1}, u_{n + 2}, \ldots)]$. In analogous to the derivation of $P'(A_n) = p_n$, we have (the summation extends over $(u_1^*, \ldots, u_n^*, u_{n + 1}, \ldots)$ such that $P[X_i = u_i^*, 1 \leq i \leq n, X_j = u_j, j > n] > 0$):
\begin{align}
 & P'(A_1 \cap A_2 \cap \cdots \cap A_n) \\
=& \sum P[X_1 = u_1^*, \ldots, X_n = u_n^*, X_{n + 1} = u_{n + 1}, X_{n + 2} = u_{n + 2}, \ldots] \\
=& \sum p_1\cdots p_nP[X_{n + 1} = u_{n + 1}, X_{n + 2} = u_{n + 2}, \ldots] \\
\overset{*}{=}& p_1\cdots p_n \\
=& P'(A_1)P'(A_2)\cdots P'(A_n).
\end{align}
The proof of "$\overset{*}{=}$" is the same as Note 2.
This completes the proof.

Note 1: By noting $\prod (1 + a_n) = \sum \log(1 + a_n)$ (for $a_n > -1$) and $\log(1 + a_n) \sim a_n$ when $a_n = o(1)$, it is easy to verify the following theorem:

If $a_n > 0$ or $a_n < 0$ for all but finitely many $n$, then the convergence status of the infinite product $\prod_n (1 + a_n)$ is the same as that of the infinite series $\sum_n a_n$.

Note 2: Proof of "$\overset{*2}{=}$":  This is equivalent to prove
\begin{align}
1 = \sum P[X_1 = u_1, \ldots, X_{n - 1} = u_{n - 1}, X_{n + 1} = u_{n + 1}, \ldots] =: P(B_n), 
\end{align}
where $B_n$ is the union of events $[X_1 = u_1, \ldots, X_{n - 1} = u_{n - 1}, X_{n + 1} = u_{n + 1}, \ldots]$ with positive probabilities.  Clearly, $B_n \supseteq \bigcup\limits_{k > n}\bigcap\limits_{m \geq k}[X_m = 1] =: E$, hence $E^c = \bigcap\limits_{k > n}\bigcup\limits_{m \geq k}[X_m = 0]$, and for every $k > n$,
\begin{align}
P(E^c) \leq P\left[\bigcup_{m \geq k}[X_m = 0]\right] \leq 
\sum_{m \geq k}P[X_m = 0] = \sum_{m \geq k}\alpha_m,
\end{align}
which implies $P(E^c) = 0$ by $\sum \alpha_n < \infty$.  Therefore, $P(B_n) \geq P(E) = 1 - P(E^c) = 1$, as desired.
A: Putting the idea of @MikeEarnest into equation:
Assume wlog $\sum_n \min(p_n,1-p_n) = \sum_n p_n$ by relabeling $p_n \rightarrow 1-p_n$ if $p_n > 0.5$.
Assume wlog let $1-p_1 < 1-p_2 <....<1-p_n < 1-p_{n+1}<...$. This is true since $\sum_n p_n < \infty$ and hence $p_n \rightarrow 0$.
Let $S$ be a continuous random variable taking value uniformly between $[0,1]$ such that $P(S \leq s) = s$ (lebesgue measure).
Let $\Omega = \{0,1,2,3...\}$.
Let $Y$ be a random variable on $\Omega$ with
$\{Y = j\} = \{1-p_j \leq S < 1-p_{j+1}\}$ for $j>0$.
$P(Y \geq j) = P(S \geq 1-p_j) = 1-P(S < 1-p_j) = p_j$.
So $A_j = \{ Y \geq j\}$ with $P(A_j) = p_j$.
I could not yet come up with events $A_j$ such that they are pairwise independent. But here is the problem with coming up with independent events.
Assume independent events $A_i$ exists.
Let $X_i = 1$ if event $A_i$ happens and $X_i = 0$ if event $A_i$ does not happen. Now $\{X_i\}$ are independent random variables. I have a conjecture, $|\Omega| > 2^{\text{Number of independent events based on $\Omega$}}$ for a discrete probability space $\Omega$. I dont know if this is true, this could be false. But something along these lines must be true. If this conjecture is true then it is impossible for countably infinite many independent events to exist in a discrete probability space $\Omega$.
