Mistake in Billingsleys book? This is about an exercise in Billingsley's book Probability and measure.
Exercise 2.15: On the field $\mathscr B_0$ in $(0,1]$ define $P(A)$ to be $1$ or $0$ according as there does or does not exist some positive $\epsilon_A$ (depending on $A$) such that $A$ contains the interval $(\frac{1}{2},\frac{1}{2} + \epsilon_A]$. Show that $P$ is finitely but not countably additive.
I am able to prove that $P$ is not $\sigma$-additve. If it were, then it would be continuous from above. Consider for example the sequence $A_n :=(\frac{1}{2}, \frac{1}{2} + \frac{1}{n}]$. $A_n \downarrow \emptyset$, but $P(A_n) \to 1 \neq 
 0 = P(\emptyset)$, since $P(A_n)=1$ for all $n$.
But I am not able to prove that $P$ is finitely additive. I believe that is, because $P$ isn't finitely additive. For example consider the sets $A = (0,1] \cap \mathbb Q$ and $B = (0,1] \backslash \mathbb Q$. They are disjoint, $P(A) = P(B) = 0$. But $A\cup B = (0,1]$ and thus $P(A \cup B)= 1 \neq P(A) + P(B)$. 
So my question is, did I make some stupid mistake? Or is there indeed a mistake in the exercise?

Your mistake: Recall that $\mathscr{B}_0$ is defined as the set of "finite disjoint unions of intervals in $(0,1]$." The issue with your counterexample is simply that the $A,B$ you use do not belong to $\mathscr{B}_0$.

Now, take any two disjoint $A,B\in\mathscr{B}_0$: by assumption, there exist $n,m\geq 1$ and disjoint non-empty intervals $I_1,\dots,I_n,J_1,\dots,J_m\subseteq (0,1]$ such that
$$
A = \bigcup_{i=1}^n I_i \, \qquad B = \bigcup_{i=1}^m J_i 
$$
We have the following cases cases:


*

*Clearly, if $P(A)=1$ and $P(B)=0$, then $P(A\cup B)=1$. Similarly if $P(A)=0$ and $P(B)=1$.

*One cannot have $P(A)=P(B)= 1$, since then by definition $1/2\in A\cap B$; but $A,B$ are taken disjoint.

*Suppose $P(A)=P(B)=0$, and by contradiction assume $P(A\cup B)=1$. This means $1/2\in A\cup B$, so wlog assume $1/2\in A$, say $1/2\in I_1$ (again wlog). Since $P(A\cup B)=1$, there exists $\varepsilon >0$ such that $(1/2, 1/2+\varepsilon] \subseteq A\cup B$: consider $(1/2, 1/2+\varepsilon] \cap I_1$. It's a non-empty intersection of intervals, so it's a non-empty interval. It is immediate to see it's of the form $(1/2, 1/2+\varepsilon']$, and it's contained in $A$: so $P(A)=1$, contradiction.
Therefore, in all possible cases we have $P(A)+P(B)=P(A\cup B)$. This shows $P$ is finitely additive.
