When can we extend a function on a $\lambda$-system to a probability measure? 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, 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?
 A: 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.
A: $\newcommand\L{\mathcal L}\newcommand\Om{\Omega}$This is to complement the nice idea by Pietro Majer by considering "choose arbitrarily a value $P_0(A)\in[0,1]$" and "this has little chance" in detail.
Let us refer to a subset $A$ of the set $\Om:=[10]:=\{1,\dots,10\}$ of cardinality $5$ as a $5$-set. Say that a $5$-set $A$ is good if $1\in A$, and say that a $5$-set $A$ is bad if $1\notin A$.
We assign the same value $p:=P_0(A)$ to all good $5$-sets and the same value $1-p$ to all bad $5$-sets.
Suppose that there is a probability measure $P$ on $2^\Om$ whose restriction to $\L:=\{\emptyset,\Om\}\cup\{A\subset\Om\colon|A|=5\}$ is $P_0$. Let $p_j:=P(\{j\})$ for $j\in\Om$.
Then $p_1+p_2+p_4+p_5+p_6=p=p_1+p_3+p_4+p_5+p_6$, so that $p_2=p_3$. Similarly, $p_2=\cdots=p_{10}$, so that $$p_2=\cdots=p_{10}=\frac{1-p_1}9.\tag{1}\label{1} $$
So,
$$p=p_1+p_2+p_3+p_4+p_5=p_1+4\dfrac{1-p_1}9=\dfrac{5p_1}9+\frac49\ge\frac49. \tag{2}\label{2} $$
Choosing now any $p\in[0,\frac49)$, we get a contradiction. $\quad\Box$
(This consideration also shows that, if we choose any $p\in[\frac49,1]$, then there will exist a probability measure $P$ on $2^\Om$ whose restriction to $\L$ is $P_0$. Indeed, in view of \eqref{2}, this probability measure $P$ is determined by the conditions $p_1=(p-\frac49)/\frac59$ and \eqref{1}.)
