$\newcommand{\scrF}{\mathcal{F}}$ Some definitions first.
Defn: Let $(\Omega, \scrF)$ be a measurable space. A signed measure on $(\Omega, \scrF)$ is a set-theoretic function $\mu:\scrF \rightarrow (-\infty, \infty]$ satsifying
- $\mu(\varnothing) = 0$.
- For any sequence of disjoint sets $\{E_n\}_{n=1}^\infty$ in $\scrF$ we have that $$ \mu\left( \bigcup_{n=1}^\infty E_n \right) = \sum_{n=1}^\infty \mu(E_n) = \lim_{N \rightarrow \infty} \sum_{n=1}^N \mu(E_n). $$ Prop: Let $(E_n)$ be a sequence in $\scrF$ and $\mu$ a measure (i.e positive) in some measure space $(\Omega, \scrF)$. Then $$ \mu\left( \bigcup_{n=1}^\infty E_n \right) \leq \sum_{n=1}^\infty \mu(E_n). $$
Both definitions made sense at first glance, but I guess after mulling it over I wasn't sure why we had to restrict ourselves to disjoint sets when $\mu$ is a signed measure. I get that it must be something to do with the fact that $\mu$ can assign negative weights to elements of $\scrF$, but is that all? Moreover what's stopping a signed measure from being sub-additive; am I not thinking abstractly enough? Because even in the case where you think of a Venn-diagram, why would $A \cup B$ every be assigned less measure than $\mu(A) +\mu(B)$ or am I just not acknowledging that the measure can be defined however you want.
I suppose I'm looking for intuitive understanding of the requirements here if one can be made.
Edit: I think a valid proof of the proposition is as follows.
To get to the sum we will construct a disjoint sequence whose union is the same as the union of the $E_n$. Define a sequence $(A_n)$ by $A_1 = E_1$ and $A_{k+1} = E_{k+1} \setminus (E_1 \cup \dots \cup E_k)$ for all $k \geq 1$. First note that these sets are disjoint; if we take some arbitrary $m$ then $A_{m}$ is disjoint from all $A_1,\dots,A_{m-1}$ and since this $m$ is arbitrary it follows that all the sets are disjoint. Moreover we have that $\cup_{i=1}^n A_i = \cup_{i=1}^n E_i$ for all $n \in \mathbb{N}$; note that the base case is satisfied trivially, so assume that $\cup_{i=1}^k A_i = \cup_{i=1}^kE_i$ for some $k \geq 0$, then \begin{align*} \bigcup_{i=1}^k A_k & = E_1 \cup \dots \cup E_k \\ \bigcup_{i=1}^{k+1} A_k & = (E_1 \cup \dots \cup E_k) \cup A_{k+1} \\ & = (E_1 \cup \dots \cup E_k) \cup (E_{k+1} \setminus (E_1 \cup \dots \cup E_k)) \\ & = E_1 \cup \dots \cup E_k \cup E_{k+1}. \end{align*} Therefore we have that $\cup_{n=1}^\infty A_n = \cup_{n=1}^\infty E_n$ and we can finish the proof. By countable additivity we can write \begin{align*} \mu \left(\bigcup_{n=1}^\infty E_n\right) & =\mu \left(\bigcup_{n=1}^\infty A_n\right) \\ & = \sum_{n=1}^\infty \mu(A_n) \\ & \leq \sum_{n=1}^\infty \mu(E_n). \end{align*}