# Borel measure of half-open and open intervals

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

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Sure. Singletons have measure $0$. – André Nicolas Sep 14 '12 at 23:13
If I'm not mistaken all intervals have the vanilla measure of b-a. – emka Sep 14 '12 at 23:34
@AndréNicolas We are supposed to prove it. – Makoto Kato Sep 15 '12 at 15:42
@MakotoKato: You gave a proof. Or else one can prove in essentially the same way that singletons have measure $0$ and conclude result from finite additivity. – André Nicolas Sep 15 '12 at 15:50

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 $\mu((a, b)) = b - a$