Coin tossing and Kolmogorov theorem Consider the space of infinite coin tossing, with $P(1)=P(0)=1/2$, $\prod_{n=1}^\infty \{0,1\}$ let $w\in\prod_{n=1}^\infty \{0,1\}$ be arbitray chosen, and $x_k(w)=a_k$ then the set $A=\{w:\sum_n x_n(w)/n<\infty\}$ has probability $0$,
I proved that $A$ is a tail event and by Kolmogorov theorem $P(A)=0$ or $P(A)=1$, I don't know, how to prove that it must be $0$
 A: I suppose that $(X_k)$ is the coordinate process.
Let $C$ be the complement of $A$ in the space $\Omega =\prod_{\Bbb N}\Omega_0$, where $\Omega_0=\{0,1\}$, considered with the uniform probability $P$ on the two atoms.
Then the "switch${}_0$" map $\Omega_0\to\Omega_0$ given by $0\leftrightarrow 1$, $a\to s_0(a):=1-a$, induces a "switch" map on $\Omega$, each component is "bit inverted". We denote by $s$ this map on "words" $w$, $w\to s(w)$.
This map is preserving the mass (the probability), since a typical cylinder (from the obvious $\sigma$-generating family) of the shape
$$
\{(a_1,\dots,a_N)\}\times \prod_{\substack{n\in\Bbb N\\n>N}}
$$
with mass $1/2^N$ is mapped bijectively to a similar cylinder with same mass.
Now let us observe that $C$ and $s(C)$ have same probability and cover $\Omega$.
Else there is some $w\in\Omega$, so that both series converge, 
$$
\begin{aligned}
\sum_{n\ge 1}\frac  1n X_n(w)&<\infty\ ,\\
\sum_{n\ge 1}\frac  1n X_n(s(w))&<\infty\ ,\qquad\text{ so their sum is also finite}\\
\sum_{n\ge 1}\frac  1n \underbrace{(X_n(w)+X_n(s(w)))}_{=1}&<\infty\ ,\qquad\text{ contradiction.}
\end{aligned}
$$
So $C$ is not a zero-set, so $A$ is a zero set.
$\square$
