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From earlier question here. Consider

$$\mu \left( (0,1) \cap \mathbb Q \right) = 0$$

where $$(0,1) = \left(0,\frac{1}{3}\right)\cup \left[\frac{1}{3}, \frac{2}{3}\right] \cup \left(\frac{2}{3}, 1\right)$$ so

$$ \mu ((0,1) \cap \mathbb Q ) = \mu \left(\left(0,\frac{1}{3}\right) \cap \mathbb Q \right) + \mu\left( \left[\frac{1}{3}, \frac{2}{3}\right] \cap \mathbb Q\right) + \mu\left(\left(\frac{2}{3}, 1\right) \cap \mathbb Q\right) = 0 + \frac{1}{3} + 0 \neq 0$$

contradiction with latter case. Why did my partition change the result?

[solved]

$\mu\left(\left[\frac{1}{3}, \frac{2}{3}\right] \cap \mathbb Q\right) = 0$.

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I think there's something you've gravely misunderstood. $\mu((0,1/3)) = 1/3$ and not $0$. –  kahen Sep 10 '11 at 16:38
    
@kahen: $(0, \frac{1}{3}) = \mathbb Q \cap (0, \frac{1}{3}) = 0$ with respect to $\mathbb R$. –  hhh Sep 10 '11 at 18:38
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When you write "$(a,b)$" in the sense of an interval, it always means interval in the reals unless you have specified otherwise. And you didn't. As for your question --- your problem is that the rationals are not Jordan measurable. It's a good exercise to prove that a bounded subset of the reals is Jordan measurable if and only if its characteristic function is Riemann integrable (or equivalently Darboux integrable). And characteristic functions of rationals in intervals aren't Riemann integrable because they're everywhere discontinuous (in said interval). –  kahen Sep 10 '11 at 18:40
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[Continued] Terry Tao wrote a good introduction to "The Problem of Measure": terrytao.wordpress.com/2010/09/04/… –  kahen Sep 10 '11 at 18:44
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