A function from an algebra to $[0, 1]$ is finitely additive iff every restriction to a finite subalgebra is finitely additive?

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: $$P(A_1\cup\cdots\cup A_n)=(P|_{\mathcal F'})(A_1\cup\cdots\cup A_n)=\sum_{i=1}^n(P|_{\mathcal F'})(A_i)=\sum_{i=1}^nP(A_i)$$