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

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 $\Lambda_N h = \Lambda_n h$

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 $$Q$$, $$2^{-nk}\sum_{y\in P_n\cap Q}g (y)=2^{-nk}g (x)\sum_{y\in P_n\cap Q}1 =2^{-nk}2^{(n-N)k}g (x)=2^{-Nk}g (x).$$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}