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So I had two guesses:

  1. The Lebesgue measure of all Borel subsets of $\mathbb{R}$ that are contained in $I$
  2. Take the subspace topology of $I$ inherited from $\mathbb{R}$. Generated a Borel $\sigma$-algebra $B(I)$ on $I$. Then each $A\in B(I)$ is a Borel set of $\mathbb{R}$ so take the Lebesgue measure of $A$.

The difference between these two guesses is that a Borel set of $\mathbb{R}$ that is also contained in $I$ may probably not in $B(I)$.

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    It feels very strange that 1 and 2 should be different..., an example of this would be nice!2012-10-28
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    I think in general I could ask 2012-10-28
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    So... you found the answer, didn't you?2012-10-29
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    The answer is that they are the same. See a post on [a more general question][1]. [1]: http://math.stackexchange.com/a/222895/266552012-10-29

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