Does such a finitely additive function exist?

Question

I was reading up about Lebesgue meaures on $\mathbb{R}$, and came across the fact that there is a Lebesgue measure $\mu$ uniquely defined by $\mu((a,b))=b-a$ on the real line. I'm trying to build up a (somewhat analogous?) idea on $\mathbb{Q}$.

I take $\mathcal{R}$ to be the set ring of subsets of $\mathbb{Q}$ which are finite unions of half-open intervals on the rational line of form $(a,b]=\{q\in\mathbb{Q}\mid a<q\leq b\}$ for $a\leq b$ rationals.

Does there then exist a finitely additive $\mu\colon\mathcal{R}\to[0,+\infty]$ such that $\mu((a,b])=b-a$ when $a\leq b$ are rationals? And if such a $\mu$ does exist, is it unique?

Thanks!

Answer 1

Yes it does exist, and is just the same as defining the Lebesgue measure on the ring $\mathcal{S}$ of subsets of $\mathbb{R}$ formed by finite unions of sets $(a,b]=\{x\in\mathbb{R}\colon a < x\le b\}$ for $a < b\in\mathbb{Q}$.

We can define $f\colon\mathcal{S}\to\mathcal{R}$ by $f(A)=A\cap\mathbb{Q}$. This is just the preimage of $A$ under the embedding $\mathbb{Q}\hookrightarrow\mathbb{R}$, and preserves set unions and differences. Then, $f$ is clearly onto and is one-to-one: if $f(A)=f(B)$ then $f(A\setminus B)=f(B\setminus A)=\emptyset$ from which we get $A\setminus B=B\setminus A=\emptyset$ and, hence, $A=B$. So, $f$ has an inverse $f^{-1}\colon\mathcal{R}\to\mathcal{S}$.

If $\nu\colon\mathcal{S}\to[0,\infty]$ is any finitely additive function, such as the restriction of the Lebesgue measure to $\mathcal{S}$ (for your case), then you can define the finitely additive function $\mu=\nu\circ f^{-1}$ on $\mathcal{R}$.

Any such additive function is uniquely defined by its values on the half-open intervals. Just apply additivity to the fact that every element of $\mathcal{R}$ (and $\mathcal{S}$) is a disjoint union of such half-open intervals.