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Let $ \displaystyle{ \{ (a_n , b_n) : n \in \mathbb N \} }$ a sequence of open intervals on $\mathbb R$ such that $ \displaystyle{[0,15] \subset \bigcup_{n=1}^{n_0} (a_n ,b_n) }$ for some $ n_0 \in \mathbb N$

Prove that $ \displaystyle{ \sum_{n=1}^{n_0} \mu(a_n ,b_n) = \sum_{n=1}^{n_0} (b_n -a_n) >15 } \quad$ where $ \mu$ is the Lebesgue measure on $ \mathbb R$

I need to prove this using induction but I can't see how.

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    O.K Thank you very much all for your answers!2012-07-03

3 Answers 3

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Use monotonicity and countable sub-additivity of $\mu$ $15=\mu ([0,15])<\mu (\bigcup_{n=1}^{n_0}(a_n,b_n))\leq\sum_{n=1}^{n_0}\mu(a_n,b_n)$ The first inequality is strict because $[0,15]$ is closed and $\bigcup_{n=1}^{n_0}(a_n,b_n)$ is open and the containment is strict.

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Without words: $ 15=15-0 < \sum_{n=1}^{n_0} (b_n-a_n). $ Why? Since $15$ is the measure of $[0,15]$, and the intervals are possibly overlapping.

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Start in induction for n0 = 1, when it is 0 there is no sense.

Use the fact that in this case Lebesgue measure is the same as the length of the interval.