# Restriction of measure to rationals

Let $X = [0,1]$ and $\mathbb Q$ - the set of rational numbers. We take $X' = X\cap \mathbb Q$ and define a measure on it such that $\lambda(X'\cap (a,b)) = b-a$ for any $a,b\in X$.

This measure is characterized by its values on atoms since there are a countable number of elements of $X'$. It's easy to see that this measure is non-unique - but can you give at least one example of such a measure $\lambda$ on rational numbers in $[0,1]$?

With an example I mean a function $p:X'\to [0,1]$ such that for any subset $A\subseteq X'$ holds $$\lambda(A) = \sum\limits_{x\in A}p(x).$$

-
Your condition $\lambda(X' \cap (a,b)) = b - a$ implies that $p(x) = 0$ for all $x$ rational by taking $a_n \nearrow x$ and $b_n \searrow x$. – t.b. Jul 23 '11 at 9:16
When working with countable sets, finitely additive measures seem to be more interesting than countably additive. (This follows from your observation, that countably additive measure is uniquely determined by measures of singletons.) – Martin Sleziak Jul 23 '11 at 9:25

## 1 Answer

There is no such measure.

As you've said yourself, since the set $\mathbb{Q} \cap [0,1]$ is countable, to give a measure is equivalent to just assigning to each point $x$ a non-negative mass $m_x$. But the condition that $$\mu( \mathbb{Q} \cap (a,b)) = (b-a) \ \text{for all} \ a< b$$ forces $m_x = 0$ for all $x$: e.g. for any positive integer $n$, $m_x \leq \mu( \mathbb{Q} \cap (x-\frac{1}{2n},x+\frac{1}{2n})) = \frac{1}{n}$.

-
Dear Pete, I have converted one of your equations to the displaystyle mode. Hope that isn't a problem. – user9413 Jul 23 '11 at 9:25