When can we extend a countably-additive set function on a Dynkin system to be a (probability) measure? Let $X$ be an arbitrary set. A class of subsets $\mathcal{A}$ of $X$ is called a $\lambda$-system (or Dynkin system) if
(1) $X\in\mathcal{A}$
(2) $A\in\mathcal{A}\Rightarrow A^c\in\mathcal{A}$
(3) $A_1,A_2,\cdots,\in\mathcal{A}$ and $A_n\cap A_m=\phi\ \forall\ m\neq n$ imply $\cup_nA_n\in\mathcal{A}$  
Suppose $\mu:\mathcal{A}\rightarrow [0,1]$ satisfies
(a) $\mu(X)=1$ and $\mu(\phi)=0$.
(b) $\mu(\cup_nA_n)=\sum_n\mu(A_n)$ if $A_n\cap A_m=\phi$ for all $n\neq m$  
Is it true that $\mu$ can be extended to a probability measure (or just finitely additive probability measure) on the sigma algebra generated by $\mathcal{A}$? That is, does there exist a probability measure $p$ on $(X,\sigma(\mathcal{A}))$ such that $p(A)=\mu(A)\ \forall\ A\in\mathcal{A}$?  
If not, are there sufficient conditions ensuring the existence of such extension?
 A: Here is an example where $\mu$ cannot be extended to a measure on $\sigma(A)$.
Let $X=\{1,2,3,4\}$,
$\mathcal A=\{\emptyset, \{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}, X\}$ is a dynkin sysmtem on $X$.
$\sigma(\mathcal A)$ is the set of all subsets of $X$.
$\mu(\emptyset)=0$,
$\mu(\{1,2\})=\mu(\{1,3\})=\mu(\{1,4\})=1/4$,
$\mu(\{3,4\})=\mu(\{2,4\})=\mu(\{2,3\})=3/4$,
$\mu(X)=1$.
$\mu$ is countably additive.
Suppose $\mu$ can be extended to $\sigma(\mathcal A)$. Then
$$\begin{aligned}
1&=\mu(X)=\mu(\{1,2,3,4\})\\
&=\mu(\{1,2\}) + \mu(\{3\}) + \mu(\{4\})\\
&\le\mu(\{1,2\}) + \mu(\{1,3\}) + \mu(\{1,4\})\\
&=\frac14+\frac14+\frac14=\frac34,
\end{aligned}$$
which is absurd.
The example is adapted from here.

To enable $\mu$ so that $\mu$ can be extended to a probability measure, we can require $\mathcal A$ be closed under binary intersection or require $\mathcal A$ be closed under binary union. Then $\mathcal A$ is, in fact, a ring of sets and $\mu$ is a finite (and, of course,  $\sigma$-finite) pre-measure on $\mathcal A$. By Carathéodory's extension theorem, $\mu$ can be extended to a measure on $\sigma(\mathcal A)$.
