Let $B$ be a box in $\mathbb{R}^d$, and let $B_1, \dots B_N$ be a collection of boxes that partition $B$. I have to demonstrate that $$|B| = |B_1| + \dots + |B_n|$$ where $|B|$ is the volume of $B$, defined in a usual way (problem from Terence Tao's Measure Theory). This seems quite obvious, but I am struggling to demonstrate this rigorously without relying on intuition. I am also not allowed to use the discretization, i.e., arguments involving approximations through limits.
My attempts:
- I considered the case $d=1$. Then $B = (a_1,b_1) \cup \dots \cup (a_n,b_n)$, and $B=(a,b)$ for some $a,b \in \mathbb{R}$. Without loss of generality we can assume that the $2n$ endpoints of these intervals are placed in the increasing order. Obviously, $a = \min(a_i) = a_1$ and $b = \max(b_i) = b_n$. Then we must have $b_1 = a_2 = \dots = b_{n-1} = a_n$ (otherwise we would have $x\notin (b_{i-1}, a_i)$ - a contradiction to $x\in (a,b)$. Summing up the lengthes of these intervals, we are left with $b_n - a_1 = b - a$ by telescoping. However, I had troubles extending this to higher dimensions.
- Another approach: intuitively corresponds to extending the borders of the boxes until the edge of the box
a
We mix all points $a_1^1, \dots a_1^N, b_1^1, \dots, b_1^N$ until $a_d^1, \dots a_d^N, b_d^1, \dots, b_d^N$ into ordered sequences $z_1^1, \dots z_1^{2N}$ until $z_d^1, \dots z_d^{2N}$. Thus the boxes are (I am being flexible with boundaries) $$ \{ (z_1^j, z_1^{j+1}) \times \dots \times (z_d^j, z_d^{j+1}) : 1\leq j < 2N\}$$ Ten any box $B_i$ can be represented as union of some boxes above. I am struggling to procede from here on.