Finite additive measure

Question

Problem: Let $[0,1]\cap\mathbb{Q} $ denote the set of all rational number inside the interval $\left[0,1\right]$, let $\mathcal{A}$ be the algebra of sets that can be expressed as finite unions of non-intersecting sets $A$ of the form $\left\{ r:a<r<b\right\} ,\left\{ r:a\leq r<b\right\} ,\left\{ r:a<r\leq b\right\} ,\left\{ r:a\leq r\leq b\right\} $, and let $\mathbb{P}\left(A\right)=b-a$. Prove that the set function $\mathbb{P}\left(A\right),A\in\mathcal{A}$, is finitely additive but not countably additive.

Attempt I have managed to show that $\mathbb{P}$ is not countably additive. However, I don't know how to show the finite additivity. Namely, if we have two disjoint sets $(a,b)$ and $(c,d)$ where $b\neq c$, how to show that $$\mathbb{P}((a,b)\cup(c,d))=\mathbb{P}(a,b)+\mathbb{P}(c,d)$$ I think if $b=c$ then everything works out fine. But this is not the case. My idea is that we have \begin{align*} \mathbb{P}((a,b)\cup[b,c]\cup(c,d))& =\mathbb{P}(a,d)\\ & = d-a\\ & = (d-c)+(c-b)+(b-a)\\ & = \mathbb{P}(a,b)+\mathbb{P}[b,c]+\mathbb{P}(c,d) \end{align*} subtract both sides by $\mathbb{P}([b,c])$, we have \begin{align*} \mathbb{P}((a,b)\cup[b,c]\cup(c,d))-\mathbb{P}([b,c])=\mathbb{P}(a,b)+\mathbb{P}(c,d) \end{align*} So it suffices to show that the left hand side is $\mathbb{P}((a,b)\cup(c,d))$.This is where I got stuck.

Answer 1

The function $\mathbb{P}$ is finitely additive by definition. It is defined by $\mathbb{P}(A) = b-a$ for any interval with endpoints $a\leq b$, and $\mathbb{P}(A_1 \cup \cdots \cup A_n) = \sum_{i=1}^n \mathbb{P}(A_i)$ for any collection of disjoint intervals $A_1,\ldots,A_n$.

What is slightly less clear is whether $\mathbb{P}$ is well-defined. For example, as you mention, it could happen that a single set can be written as the finite disjoint union of intervals in several ways, for example $[a,b] = [a,c) \cup [c,b]$. In this particular case, there is no problem, since using the decomposition on the left we get a measure of $b-a$, and using the decomposition on the right we get a measure of $(c-a) + (b-c) = b-a$. But how do we know that this is always the case? This is exactly the point of the exercise.

You can probably prove that $\mathbb{P}$ is well-defined from first principles, but a shorter route is to show that $\mathbb{P}$ agrees with the standard Lebesgue measure on its domain of definition.