Help understanding an inequality on Rudin's construction of the Lebesgue measure

Question

I am having trouble understanding an inequality in Theorem 2.20 from "Real and Complex Analysis."

Rudin states that if $f\in\operatorname{C}_c(\mathbb{R}^k)$ , $f$ is real, $W$ is an open k-cell which contains the support of $f$, and $\epsilon>0$ , such that

(i) $g$ and $h$ are constant on each box belonging to $\Omega_N$

(ii) $g\leq f\leq h$

(iii) $h-g<\epsilon$

If $n>N$, Property 2.19(c) shows that

$$\Lambda_N g = \Lambda_n g \leq \Lambda_n f \leq \Lambda_n h = \Lambda_N h$$

Here $\Lambda$ is defined as

$$\Lambda_n f := \lim\limits_{n \to \infty} 2^{-nk} \sum\limits_{x \in P_n} f(x)$$

$\Omega_n$ is

the collection of all $2^{-n}$ boxes with corners at $P_n$

$P_n$ is

the set of all $x\in\mathbb{R^k}$ whose coordinates are integral multiples of $2^{-n}$

Property 2.19 (c) is

For $\{\Omega_n\}$, if $Q\in \Omega_r$, then vol$(Q)=2^{-rk}$; and if $n>r$, the set $P_n$ has exactly $2^{(n-r)k}$ points in $Q$

What I don't understand is how Property 2.19(c) implies that $\Lambda_N g = \Lambda_n g$ and $\Lambda_N h = \Lambda_n h$

Answer 1

When you calculate $\Lambda_ng $, you are adding over more boxes, i.e, you are subdividing the boxes from $\Omega_n $. But $g $ is constant over those additional boxes. Concretely, a fixed $x\in P_N $ corresponds to some box $Q\in\Omega_N $; then, as $g $ is constant in $Q $, $$ 2^{-nk}\sum_{y\in P_n\cap Q}g (y)=2^{-nk}g (x)\sum_{y\in P_n\cap Q}1 =2^{-nk}2^{(n-N)k}g (x)=2^{-Nk}g (x). $$