Ex3-7

Chapter 3 Question 7
Lemma 3.15: If \{ I_{k} \}_{k=1}^{N} is a finite collection of nonoverlapping intervals in \mathbf{R}^n, then \bigcup_{k=1}^{N} I_{k} is measurable and \mu(\bigcup_{k=1}^{N} I_{k}) = \sum_{k=1}^{N} \mu(I_{k}) .
Proof: Corollary 3.13 tells us that if I is an interval, then I is measurable and \mu(I) = v(I). Since a finite collection of sets is clearly countable, and each set I_{k} is measurable, Theorem 3.12 tells us that the union \bigcup_{k=1}^{N} I_{k} is measurable. We are left to show that the equality \mu(\bigcup_{k=1}^{N} I_{k}) = \sum_{k=1}^{N} \mu(I_{k}) holds. 
Note that since \bigcup_{k=1}^{N} I_{k} is measurable, its Lebesgue measure must agree with its outer measure. Thus it suffices to calculate the outer measure of \bigcup_{k=1}^{N} I_{k}. By Theorem 3.4, we have the bound 
|\bigcup_{k=1}^{N} I_{k})|_{e} \leq \sum_{k=1}^{N} |I_{k}|_{e}. 
Note also that |I_{k}|_{e} = v(I_{k}) for each k, so we can rewrite the above as 
|\bigcup_{k=1}^{N} I_{k})|_{e} \leq \sum_{k=1}^{N} v(I_{k}) 
Thus we are reduced to the problem of showing the other direction, namely 
|\bigcup_{k=1}^{N} I_{k})|_{e} \geq \sum_{k=1}^{N} v(I_{k})
We show the above by using compactness. Let S=\bigcup_{l}J_l be a covering of \bigcup_{k=1}^{N} I_{k} by intervals. Given an arbitrary \epsilon > 0, let J^{*}_{l} contain J_{l} in its interior such that 
v(J^{*}_{l}) \leq (1 + \epsilon) v(J_{l}) (*)
Then we have I_{k} \subset \bigcup_{l}(J^{*}_l)^{\circ} for every k. Since \bigcup_{k=1}^{N} I_{k} is clearly compact (it is closed and bounded) in the Euclidean topology, by Heine-Borel we can extract a finite subcover \bigcup_{l=1}^{M}J^{*}_l of \bigcup_{k=1}^{N} I_k. 
By a basic fact about the combinatorics of intervals in \mathbf{R}^n which is intuitive but tedious to prove, and because \{J_l^*\cap I_k\}_{1\leq l\leq M} is a cover for I_k, we have
v(I_k) \leq \sum_{l=1}^{M}v(J^{*}_l\cap I_k)
Summing the inequalities over k, we get
\sum_{k=1}^{N} v(I_k) \leq \sum_{k=1}^N\sum_{l=1}^{M}v(J^{*}_l\cap I_k) = \sum_{l=1}^M \sum_{k=1}^N v(J^*_l \cap I_k)\leq \sum_{l=1}^M v(J_l^*)
since the collection \{I_k\}_{1\leq k\leq N} is nonoverlapping. 
Using (*), we then have 
\sum_{k}^{N} v(I_k) \leq \sum_{l=1}^{M}v(J^{*}_l) \leq (1 + \epsilon) \sum_{l=1}^{M}v(J_l)
Since \epsilon > 0 was chosen arbitrarily, we conclude 
\sum_{k}^{N} v(I_k) \leq \sum_{l=1}^{M}v(J_l)
Since we also have \sum_{l=1}^{M}v(J_l) \leq \sigma(S), we can conclude that
\sum_{k}^{N} v(I_k) \leq \sigma(S)
and so by taking infimums on both sides, taken over all possible interval coverings, we arrive at
\sum_{k}^{N} v(I_k) \leq |\bigcup_{k} I_{k})|_{e}
as desired.''
Comment: I used \mu throughout to denote the Lebesgue measure in Euclidean space, which differs from the book. I suppose somebody who is bothered by this can help me change it to the notation used in the book :).