Jordan Measure, Semi-closed sets and Partitions

222 Views Asked by At

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]

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