Won't signed measures upset Riemann?

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I can't seem to wrap my head around signed measures.

If $\mu$ is a signed measure on a measurable space $(X, \Sigma)$, then there'll be sets of both positive and negative measure. Let $E_1, E_2, \ldots\in\Sigma$ be disjoint and set $E := \bigcup_i E_i$ which is also in $\Sigma$. Now, the definition requires that $\mu(E) = \sum_{i = 1}^\infty \mu(E_i)$. However, since there can be infinitely many terms, both positive and negative, it well may be that the sequence $\sum_{i = 1}^\infty \mu(E_i)$ converges only conditionally. Then Riemann says that you can rearrange the sequence to converge to any extended real! And this would mean that $\mu(E)$ is not just one value—which is nonsensical!

Question: What am I missing here?