I think de Finetti proved the proposition in the title, namely,

If $\mathcal{F}$ is an algebra and $P : \mathcal{F} \rightarrow [0, 1]$, then $P$ is a finitely additive probability measure on $\mathcal{F}$ iff the restriction $P|_{\mathcal{F}'}$ of $P$ to $\mathcal{F}'$ is a finitely additive probability measure on $\mathcal{F}'$, for all finite subalgebras $\mathcal{F}'$ of $\mathcal{F}$.

By $P|_{\mathcal{F}'}$, I mean the function $P|_{\mathcal{F}'} : \mathcal{F}' \rightarrow [0, 1]$ such that $P|_{\mathcal{F}'}(X) = P(X)$ for all $X$ in $\mathcal{F}'$.

But I can't find the proof. So I'm looking for assurance that the proposition is true, but also a reference for the proof. Thanks!

Let $A_1,\dots,A_n\in\mathcal F$.

They induce an algebra $\mathcal F'$ in the sense that $\mathcal F'$ is the smallest algebra that contains the $A_i$. Then automatically $\mathcal F'$ is a subalgebra of $\mathcal F$.

Elements of this subalgebra are unions of sets that belong to the collection $$\{E_1\cap\cdots\cap E_n\mid E_i\in\{A_i,A_i^{\complement}\}\text{ for }i=1,\dots,n\}$$ Note that this collection contains at most $2^n$ elements, so that $\mathcal F'$ contains at most $2^{2^n}$ elements.

So $\mathcal F'$ is a finite subalgebra and if $A_1,\dots, A_n$ are disjoint then consequently: $$P(A_1\cup\cdots\cup A_n)=(P|_{\mathcal F'})(A_1\cup\cdots\cup A_n)=\sum_{i=1}^n(P|_{\mathcal F'})(A_i)=\sum_{i=1}^nP(A_i)$$

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