Let $\sum_{n=0}^\infty a_n$ be a conditionally convergent series of real numbers. Riemann's rearrangement theorem says that, for every $s\in\Bbb R$, there is some bijection $b\colon\Bbb Z_+\longrightarrow\Bbb Z_+$ such that $\sum_{n=0}^\infty a_{b(n)}=s$.
However, it is intuitive that if you take a random bijection $b\colon\Bbb Z_+\longrightarrow\Bbb Z_+$, then the series $\sum_{n=0}^\infty a_{b(n)}$ diverges. I would like to be able to state this in a formal way. To be more precise, let $\mathcal B$ be the set of all bijections from $\Bbb Z_+$ into itself.
Is there a non-trivial measure $\mu$ in the $\sigma$-algebra $\mathcal P(\mathcal B)$ such that, for every conditionally convergent series $\sum_{n=0}^\infty a_n$,$$\mu\left(\left\{b\in\mathcal B\,\middle|\,\sum_{n=0}^\infty a_{b(n)}\text{ converges}\right\}\right)=0.$$
I've put “non-trivial” in the question in order to avoid taking $\mu$ equal to the null function.
Best Answer
There is no such measure under a hypothesis known to be consistent with ZFC.
I'll assume that you want $\mu$ to be a probability measure (equivalently a $\sigma$-finite measure, as for every $\sigma$-finite measure there is a probability measure with the same null $\sigma$-ideal). It is possible to generalize the following to the case of a semifinite measure, at the cost of complicating the argument somewhat. I have also followed your convention that "conditionally convergent" means "properly conditionally convergent", i.e. there is some rearrangement that diverges.
The following set-theoretic hypothesis is consistent with ZFC.
Where point-supported means that for every set $S \subseteq \mathcal{B}$ such that $\mu(S) > 0$, there exists a point $x \in S$ such that $\mu(\{x\}) > 0$.
Proof of relative consistency of the Hypothesis: Suppose there is a $\mu$ that is not point-supported. Then there is a set $S \subseteq \mathcal{B}$ such that $\mu(S) > 0$ and for all $x \in S$, $\mu(\{x\}) = 0$. We can restrict $\mu$ to $S$, which gives us a set and a finite measure defined on the power set, vanishing on all singletons. This implies that there exists a real-valued measurable cardinal $\kappa$ with $\kappa \leq |S| \leq |\mathcal{B}| = 2^{\aleph_0}$. Ulam proved (see Jech's Set Theory Theorem 10.1, or more specifically Definition 10.11 to Corollary 10.15) that a real-valued measurable cardinal is weakly inaccessible (uncountable, not a successor, and not a strictly smaller union of strictly smaller cardinals). Therefore the Hypothesis is implied by the continuum hypothesis, or even by the continuum having any smallish sort of value like $2^{\aleph_0} = \aleph_{\omega_1}$. $\Box$
Now let's answer the question under the set-theoretic Hypothesis. We require the following proposition.
Proof of nonexistence of the required measure, from the Hypothesis and the Proposition: Let $\mu$ be a probability measure on $(\mathcal{B},\mathcal{P}(\mathcal{B}))$. Since $\mu$ is point-supported, there exists an element $b \in \mathcal{B}$ such that $\mu(\{b\}) > 0$. By the Proposition, there exists a conditionally convergent series whose sum is preserved by permuting with $b$. Therefore it is not true that for all conditionally convergent series, $\mu$ vanishes on the set of permutations preserving their convergence, for any probability measure $\mu$. $\Box$
If you don't want to prove the Proposition yourself, here is a proof. We need a lemma.
Proof of the Lemma: For all $k \in \mathbb{Z}_+$, there exists $m \in \mathbb{Z}_+$ such that $f(m) \leq k < f(m+1)$ (just take $m$ to be the largest such that $f(m) \leq k$). By the definition of $(b_n)_{n \in \mathbb{Z}_+}$: $$ \sum_{n=0}^k b_n = \sum_{n=0}^{m} b_{f(n)} + 0 = \sum_{n=0}^m a_n. $$ Define $a = \sum_{n=0}^\infty a_n$. Given $\epsilon > 0$, there exists $N \in \mathbb{Z}_+$ such that for all $m \geq N$, $\left| a - \sum_{n=0}^m a_i\right| < \epsilon$. So for all $k \geq f(N)$, there exists $m \in \mathbb{Z}_+$ such that $f(m) \leq k < f(m+1)$ and $m \geq N$, so $$ \left| a - \sum_{n=0}^k b_n \right| = \left| a - \sum_{n=0}^m a_n \right| < \epsilon, $$ proving that $(b_n)_{n \in \mathbb{Z}_+}$ is summable to the same sum $a$. If $(b_n)_{n \in \mathbb{Z}_+}$ were absolutely summable, then $(|b_n|)_{n \in \mathbb{Z}_+}$ would be summable, and by a similar argument to the above $\sum_{n=0}^{f(m)}|b_n| = \sum_{n=0}^m |a_n|$, and $(a_n)_{n \in \mathbb{Z}_+}$ would be absolutely summable, but as it is not this means that $(b_n)_{n \in \mathbb{Z}_+}$ is (properly) conditionally convergent. $\Box$
Proof of the Proposition: Let $b \in \mathcal{B}$. It is slightly simpler to find an $(a_n)_{n \in \mathbb{Z}_+}$ such that $\sum_{n=0}^\infty a_{b^{-1}(n)} = \sum_{n=0}^\infty a_n$, so we aim for this, knowing that we only need to replace $b$ with $b^{-1}$ at the beginning to get the other form. Define $f : \mathbb{Z}_+ \rightarrow \mathbb{Z}_+$ by taking $f(0) = 0$ and $f(n+1)$ to be the smallest $m$ such that m > f(n) and $b(m) > b(f(n))$. Such an $m$ exists because $\{ m \in \mathbb{Z}_+ \mid m \leq f(n) \}$ and $\{ m \in \mathbb{Z}_+ \mid b(m) \leq b(f(n)) \}$ are both finite, so their union is not all of $\mathbb{Z}_+$. Having constructed $f$, we have $f(n+1) > f(n)$ and $b(f(n+1)) > b(f(n))$ for all $n \in \mathbb{Z}_+$, so by induction $f$ and $b \circ f$ are monotone and injective.
Define $(c_n)_{n \in \mathbb{Z}_+}$ to be your favourite conditionally (and non-absolutely) convergent series, for instance $c_n = (-1)^n\frac{1}{n+1}$. Define $(a_n)_{n \in \mathbb{Z}_+}$ by $a_n = c_{f^{-1}(n)}$ for $n \in \mathrm{im}(f)$ and $a_n = 0$ otherwise. By the Lemma, $(a_n)_{n \in \mathbb{Z}_+}$ is conditionally convergent as well. We have $a_{b^{-1}(n)} = f^{-1}(b^{-1}(n))$ iff $b^{-1}(n) \in \mathrm{im}(f)$ iff $n \in \mathrm{im}(b \circ f)$, and $a_{b^{-1}(n)} = 0$ otherwise, so we can apply the Lemma again and conclude that $(a_{b^{-1}(n)})_{n \in \mathbb{Z}_+}$ is conditionally summable to the same sum as $(c_n)_{n \in \mathbb{Z}_+}$ and therefore $(a_n)_{n \in \mathbb{Z}_+}$. $\Box$
Final remark: I do not know if a positive answer is possible if $2^{\aleph_0}$ is a real-valued measurable cardinal, nor even how to answer such a question.