So I had two guesses:
- The Lebesgue measure of all Borel subsets of $\mathbb{R}$ that are contained in $I$
- 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 $B(I)$.