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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 $$ \lambda(A) = \sum\limits_{x\in A}p(x). $$

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    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 $b_n \searrow x$.2011-07-23
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    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.)2011-07-23

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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 $n$, $m_x \leq \mu( \mathbb{Q} \cap (x-\frac{1}{2n},x+\frac{1}{2n})) = \frac{1}{n}$.

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    Dear Pete, I have converted one of your equations to the displaystyle mode. Hope that isn't a problem.2011-07-23
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    Why is putting a measure on a countable space equivalent to assigning a nonnegative mass to each point? This doesn't seem true to me. can't we assign positive measure to intervals of rational numbers, while assigning zero mass to each individual rational number, just as we would with the real numbers? why not?2018-02-21
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    @Programmer2134: No, we can't, because intervals in $\mathbb{Q}$ are countable sets and measures are by definition countably additive.2018-02-21