Im using the book by Terrence Tao to learn the Lebesgue measure. I started from the beginning and I seem to not understand this very basic proof. Could you please give me an idea of what the author means? Or a more visual/geometrical/physical idea of what is happening? It goes like this:
Lemma 1.1.2 (Measure of an elementary set). Let E ⊂ $R^d$ be an elementary set.E can be expressed as the finite union of disjoint boxes.
Proof:
We first prove (i) in the one-dimensional case d = 1. Given
any finite collection of intervals $I_1, . . . , I_k$, one can place the $2k$ endpoints of these intervals in increasing order (discarding repetitions).
Looking at the open intervals between these endpoints, together with
the endpoints themselves (viewed as intervals of length zero), we see
that there exists a finite collection of disjoint intervals $J_1, . . . , J_{k'}$
such that each of the $I_1, . . . , I_k$ are a union of some subcollection of
the $J_1, . . . , J_{k'}$ . This already gives (i) when d = 1. To prove the
higher dimensional case...
For example if you have the Intervalls \begin{align*} I_1 &:= \left[-1, \frac{1}{2}\right] \\ I_2 &:= \left[\frac{1}{2}, 2\right] \\ I_3 &:= [0, 1] \end{align*} then you get the endpoints $\{-1, \frac{1}{2}, \frac{1}{2}, 2, 0, 1\} = \{ -1, 0, \frac{1}{2}, 1, 2\}$ with duplicates removed and sorted in increasing order you have the Intervalls $J$ such that \begin{align*} J_1 &:= (-1, 0) \\ J_2 &:= \left(0, \frac{1}{2}\right) \\ J_3 &:= \left(\frac{1}{2}, 1\right) \\ J_4 &:= (1, 2) \end{align*} and you can write \begin{align*} I_1 &:= J_1 \cup J_2 \\ I_2 &:= J_3 \cup J_4 \\ I_3 &:= J_2 \cup J_3 \end{align*} with those unions being disjoined.
Now for $I := I_1 \cup I_2 \cup I_3$ you can write \begin{align*} I &= J_1 \cup J_2 \cup J_3 \cup J_4 \cup J_2 \cup J_3 \cup \{-1, 0, \frac{1}{2}, 1, 2\} \\ &= J_1 \cup J_2 \cup J_3 \cup J_4 \cup \{-1, 0, \frac{1}{2}, 1, 2\} \end{align*} with those unions being disjoined.