Ex3-7

Chapter 3 Question 7

Lemma 3.15: If $$\{ I_{k} \}_{k=1}^{N}$$ is a finite collection of pairwise disjoint 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 :).