Let $\Omega$ be a nonempty set and let $\mathcal{L}$ a $\lambda$-system on $\Omega$. That is,


(i) $\Omega \in \mathcal{L}$,  

(ii) if $A, B \in \mathcal{L}$ and $A \subseteq B$, then $B \setminus A \in \mathcal{L}$, and  

(iii) if $\{A\_n\}\_{n = 1}^\infty \subseteq \mathcal{L}$ and $A\_n \subseteq A\_{n + 1}$, then $\bigcup\_n A\_n \in \mathcal{L}$.


Let $P\_0: \mathcal{L} \to [0,1]$ satisfy


(a) $P\_0(\Omega) = 1$,  

(b) if $A, B \in \mathcal{L}$ and $A \subseteq B$, then $P\_0(B \setminus A) = P\_0(B) - P\_0(A)$, and  

(c) if $\{A\_n\}\_{n = 1}^\infty \subseteq \mathcal{L}$ and $A\_n \subseteq A\_{n + 1}$, then $P\_0(\bigcup\_n A\_n) = \lim\_n P\_0(A\_n)$.


Is it necessarily the case that there exists a probability measure $P$ on $(\Omega, \sigma(\mathcal{L}))$ such that $P|\_\mathcal{L} = P\_0$?


I have not seen anything like this before and my initial search did not turn up anything. Granted, it is a little hard to search for, since most of the search results are about the $\pi$-$\lambda$ theorem. The same question [was asked 5 years ago on Math SE](https://math.stackexchange.com/q/2528391/11867), but it received no comments or answers. I started a bounty on it, but if no one answered it 5 years ago, I doubt it will gain traction. Hence, I thought it appropriate to post here. Has anyone seen anything like this or know of any references that discuss this question?



When can we extend a function on a $\lambda$-system to a probability measure?
The answer is no, for quite a trivial reason: $\mathcal{L}$ may have not enough pairs $A\subset B$, and not enough monotone increasing sequences $(A\_n)\_n$, to make $(a)$ and $(b)$ meaningful. For instance, take $\Omega:=\{1,2,\dots,10\}$ and $\mathcal{L}:=\{A\subset \Omega: |A|= 5\}\cup\{\Omega\}\cup\{\emptyset\}$. It is quite obviously a $\lambda$-system, and generates the whole power set algebra $2^\Omega$.


Define $P\_0(\Omega):=1$, and choose arbitrarily a value $P\_0(A)\in[0,1]$, for those $\frac12 {10 \choose 5}>100$ sets $A\in \mathcal L\setminus\{\Omega\}$ such that $1\in A$, and for the remaining ones, put $P\_0(A):=1-P\_0(A^c)$. Of course this has little chance to be the restriction of a measure on $2^\Omega$, which is determined by just the values on the $10$ points.


