Why does $P\{\theta_n = \pm 1\}= \frac{1}{2}$ necessarily exist? A step in the proof of Theorem 4.17 from Foundations of Modern Probability edition 2 by Olav Kallenberg says:

Let $\xi_1,\xi_2,...$ be independent, symmetric random variables, assume $\sum_n \xi_n^2 =\infty$ a.s., introduce an independent sequence of i.i.d. random variables $\theta_n$ with $P\{\theta_n = \pm 1\} =\frac{1}{2}$, and note that the sequence $(\xi_n)$ and $(\theta_n|\xi_n|)$ have the same distribution.

Why does the sequence $\theta_n$ exist? 
If there exists a measurable set $A$ s.t. $P\{A\} = \frac{1}{2}$ then it's done, but such $A$ doesn't seem necessarily exist.
If $P\{\xi_n = 0\} = 0$ for some $n$, then $P\{\xi_n>0\} = P\{\xi_n<0\}= \frac{1}{2}$ by symmetry so we can let $A$ above equal to $P\{\xi_n>0\}$ but we may have $P\{\xi_n = 0\} > 0\,\, \forall n$.
 A: You can always extend definitions of the $\ \xi_n\ $ from the probability space $\ (\Omega,\mathcal{F}, P)\ $ on which they were originally defined to (for example) the product space $\ (\Omega\times[0,1), \sigma(\mathcal{F}\times\mathcal{M}), P\times\ell)\ $, where $\ \ell\ $ is Lebesgue measure on the set, $\ \mathcal{M}\ $, of measurable subsets of $\ [0,1)\ $, by taking $\ \xi_n(\omega, x)=\xi_n(\omega)\ $ for all $\ \omega\in\Omega\ $ and $\ x\in[0,1)\ $. You can then obtain the desired $\ \theta_n\ $ as
$$
\theta_n(\omega, x)=(-1)^{\lfloor2^nx\rfloor}
$$
for $\ \omega\in\Omega\ $ and $\ x\in[0,1)\ $.
A: It is not assumed that $(\theta_n)$'s exist on the same space. It is rarely necessary to construct random variables on a particular space to prove theorems in Probability Theory. What Kallenberg wants is some probability space on which random variables $\xi_n'$'s and $\theta_n$'s exist such that $(\xi_n')$ has the same distribution as $(\xi_n)$, $(\theta_n)$ has the stated properties and $\xi_n'$'s are independent of  $\theta_n$'s. Existence of such random variables is guaranteed by Kolmogorov's Existence Theorem (also known as Kolmogorov's Consistency Theorem). 
