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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! – Per Alexandersson Oct 28 '12 at 17:01
    
I think in general I could ask <math.stackexchange.com/questions/222810/…; – Montez Oct 28 '12 at 17:15
    
So... you found the answer, didn't you? – Did Oct 29 '12 at 14:05
    
The answer is that they are the same. See a post on [a more general question][1]. [1]: math.stackexchange.com/a/222895/26655 – Montez Oct 29 '12 at 22:55

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