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Take $P : F \rightarrow R$ a set-function closed wrt countable unions and complements.

$P(\sum_1^\infty A_i) = \sum_1^\infty P(A_i)$ ($A_i \text{ disjoint from } A_j \forall i \neq j$)
is equivalent to :
$A_n \rightarrow A \implies P(A_n) \rightarrow P(A)$ (sets not necessarily disjoint)

I was thinking :
Take $H_n = A_n - H_{n-1}$ with $H_1 = A_1$ with A-B = $A \cap B^c$ then
$A_n = \cup_{1}^{n} H_i$ (which is only true with $A_i \subset A_{i+1}$), so $A = \cup_1^\infty A_n = \cup_1^\infty H_n$. Where we would use the disjointness of $H_i$ 's and the first proposition. But what would be a more general argument that would take into account all possible set convergences?
And how does the second proposition imply the first? I only manage to see the equivalence in the case where $A_i \subset A_{i+1}$. This also includes the case of $A_{i+1} \subset A_i$ by complementation.

I think that for the general case we can use $A = \cup_{n=1}^\infty \cap_{k=n}^\infty A_k$ in general and $A = \cup A_n$ when $A_i \subset A_{i+1}$ or $A = \cap A_n$ when $A_{i+1} \subset A_{i}$

Any hints and insight would be greatly appreciated!!

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Hint

Show the inequality $$ P(\lim\inf A_n)\leq \lim\inf P(A_n)\leq\lim\sup(A_n)\leq P(\lim\sup A_n) $$ and use the fact that if $A_n\to A$, then $\lim\inf A_n=\lim\sup A_n=A.$

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