the Borel set is the $\sigma$-ring generated by the open sets. One possible Borel measure on the real line is defined, for a closed interval, as:
$$\mu([a,b])=b-a$$
But, from my understanding, intervals of the type $[a,b)$, or $(a,b)$ are also part of the Borel set, as it is also generated by the compact sets.
How do you compute the measure of those half-open and open intervals ? Is it $b-a$ too ?
Thanks,
JD
Best Answer
Let $k$ be an integer such that $k > \frac{1}{b-a}$.
$[a, b - 1/n] \subset [a, b - 1/(n+1)]$ for $n \ge k$.
$[a, b) = \bigcup_{n\ge k} [a, b - 1/n]$.
Hence $\mu([a, b)) = \lim_{n\ge k} \mu([a, b - 1/n]) = b - a$. Hence, similiarly $\mu((a, b)) = b - a$