Countably additive property of Lebesgue measure I am farily new to the world of measure theory, and I am working on the Billingsey (third edition). 
At page 23 the field $B_0$ is defined as the field of finite disjoint union of subintervals of $\Omega$. However, to prove that the Lebesgue measure is countably additive probability measure on the field B0, it says (page 26)
Suppose that $A=\bigcup_{k=1}^\infty A_k$, where $A$ and $A_k$ are $B_0$-sets and the $A_k$ are disjoint. 
How can the union of infinite elementes be in a set which is defined as the set of finite unions?
Of course, this raises a more foundamental question: if $A=\bigcup_{k=1}^\infty A_k$ is in $B_0$, and we already know that $B_0$ is a field, then isn't $B_0$ also a $\sigma$-algebra, since it is closed under countable unions? The book states that this is not the case (and invests quite few pages later to extend the Lebesgue measure on a Borel set, a $\sigma$-algebra), but I don't understand why. I am certainly missing some important point here.
 A: The operative phrase here is suppose that.  In general $B_0$ will not be closed under taking countable unions; it's not a $\sigma$-algebra and nobody is claiming that it is.  But it could happen that there is some sequence of disjoint sets $A_1, A_2, \dots$ from $B_0$, such that their union $A = \bigcup_{k=1}^\infty A_k$ just happens to be another set in $B_0$.  When this happens, we want it to be the case that $m(A) = \sum_{k=1}^\infty m(A_k)$.
I'm not sure whether "subintervals" here means open or closed or what.  Let's say $B_0$ is the field of finite unions of half-open intervals in $\Omega = [0,1)$.  In general a countable union of such sets need not be a finite union of half-open intervals.  Consider something like $A_1 = [0,1/2)$, $A_2 = [3/4, 7/8)$, $A_3 = [15/16, 31/32)$, and so on.  The union $A = \bigcup_{k=1}^\infty A_k$ is definitely not in $B_0$, in this case.   But in some other cases it could be.
For example, suppose $A_1 = [0,1/2)$, $A_2 = [1/2, 3/4)$, $A_3 = [3/4, 7/8)$, and so on.  Each of the sets $A_k$ is in $B_0$, and their union is $[0,1)$ which is also in $B_0$.  We can then check that $m([0,1)) = 1 = \sum_{k=1}^\infty 2^{-k} = \sum_{k=1}^\infty m(A_k)$ so countable additivity checks out for this case.  The claim is that it holds in all such cases.
