Can we assign a measure on sequences $\epsilon_n\in\{-1,1\}$ such that $\sum_{n\ge 1}a_n \epsilon_n $ converges? If so, what is the image? Let $(a_n)_n$ be a sequence of real numbers, let $(\epsilon_n)_n$ be a sequence of i.i.d random variables with $P(\epsilon_n=1)=P(\epsilon_n=-1)=1/2$, and let $m$ denote Lebesgue measure. Consider the spaces
$$
\mathcal{A} = \{(a_n)_n: a_n\in\mathbb{R}\};\qquad \mathcal{E} = \{(\epsilon_n)_n: \epsilon_n = \pm 1\}
$$
$\mathcal{E}$ can be considered a "random choice of sign" in the probabilistic context, for example. Lastly, define a map
$$
r:\mathcal{E}\to[0,1], (\epsilon_n)_n\mapsto \sum_{n\ge1}\frac{1+\epsilon_n}{2^{n+1}}
$$
My question: is the following map
$$
f: \mathcal{A} \to \left[0,1\right],\,
(a_n)_n \mapsto m\left(\left\{x\in[0,1]:\sum_{n\ge 1}{r^{-1}(x)} a_n \text{ converges}\right\}\right)
$$well-defined? If so, is it surjective; if not, what is its image?
For example, if $(a_n)_n=1/n$, by Kolmogorov's Three-Series theorem, $\sum_{n\ge 1} \epsilon_n/n$ converges almost surely, so $f((1/n)_n)=1$; likewise, $f((1/\sqrt{n})_n)=0$.
Many sequences (in particular, those that whose series are absolutely convergent) map to $1$ and many sequences map to $0$ (such as sequences with $\lim_{n\to\infty} a_n\ne 0$), but I struggled to find a sequence $(b_n)_n$ with $0<f((b_n)_n)<1$. I've read a bit about random harmonic series and the Three-Series theorem but haven't made it much past that.
 A: Convergence of a series such as this one an example of a tail event and by the Kolmogorov Zero-One Law must have probability $0$ or $1$. This is a special case of a slightly more general result:

Suppose that $X_1,X_2,X_3,X_4,\ldots$ is a sequence of independent random variables. The probability that $\sum X_i$ converges is either $0$ or $1$.

Your question asks about the special case where we have fixed some sequence of real numbers $a_n$ and define the $X_n$ to be independent random variables taking the values $a_n$ or $-a_n$ with equal probability.
We can apply the zero-one law as follows. First, you ask if the probability is well-defined, so let's note that the event that $\sum X_i$ converges is indeed measurable. This may be observed by noting that, for real numbers $s_i$, the convergence of $\sum s_i$ can be stated in terms that translate easily to a countable intersection of countable unions of countable intersections of measurable sets:

$\sum s_i$ converges if and only if for every $\ell \in\mathbb N$ there exists some rational $p,q\in\mathbb Q$ and some $N\in\mathbb N$ such that $|p-q| < 1/\ell$ and for every $n\geq N$ we have $p <\sum_{i=1}^n s_i < q$.

There's a bit of analytical trickery here - we've replaced "convergence" by asking that the partial sums get restricted to arbitrarily small intervals eventually. Showing that convergence implies this statement requires knowing, essentially, that the open intervals with rational endpoints form a base for the topology on $\mathbb R$. The converse direction requires completeness of $\mathbb R$. This said, one may easily translate the resulting definition to operations that play well with the measure, hence the convergence of $\sum X_i$ does have a well-defined probability.
A tail event is any event that is independent of any finite collection of the sequence of random variables. Convergence is a tail event because $\sum_{i=1}^{\infty} X_i$ converges exactly when $\sum_{i=N}^{\infty} X_i$ converges for some larger $N$ - and the latter event is clearly independent of the variables $X_1,\ldots,X_{N-1}$ given that it is some condition involving only $X_N,X_{N+1},\ldots$ (which are, by assumption, each independent of the earlier values).
This suffices to show the conditions of the zero one law - hence we can immediately apply it to see that the probability of convergence can only be $0$ or $1$.
