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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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6  
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

1 Answer 1

up vote 2 down vote accepted

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

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