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?


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    $\begingroup$ @Alex: The restriction to $\mathbb Q$ is the constant function $0$. $\endgroup$
    – Asaf Karagila
    Nov 1, 2011 at 23:24
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    $\begingroup$ @AsafKaragila: I'm sorry, I suppose I'm being dense. Could you explain why? $\endgroup$ Nov 1, 2011 at 23:25
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    $\begingroup$ @Alex: The usual measure is countably-additive and is zero on singletons. Thus zero on every countable set. Ergo zero on every set of rationals. $\endgroup$
    – Asaf Karagila
    Nov 1, 2011 at 23:32
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    $\begingroup$ If $A\in\mathcal R$, write is as a finite union of disjoint semi-open intervals and define $\mu(A)$ to be the Lebesgue measure of those intervals now viewed in $\mathbb R$. $\endgroup$ Nov 1, 2011 at 23:48
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    $\begingroup$ @Yuval: That won't give you a measure that is additive on disjoint subsets. $\endgroup$ Nov 2, 2011 at 0:26

1 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.

  • $\begingroup$ Also, this is essentially the same idea as suggested by Mariano in a comment above. $\endgroup$ Nov 2, 2011 at 1:21

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