why $\Lambda_N g = \Lambda_n g $? In Rudin RCA theorem $2.20 $ Page no:$51$
Rudin say that if $n >N$, then we have
$$\Lambda_N g =\Lambda_n g \leq \Lambda_n f \leq \Lambda_n h = \Lambda_N h$$

Here I'm confused that  why $\Lambda_N g =\Lambda_n g $?

My attempt:
It is given that $g$ is constant on each box in $\{\Omega_n\}$
Now if  $n > N$ , then $$\Lambda_n g :=  2^{-nk} \sum\limits_{x \in P_n} g(x)$$
Now  using the Property :
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 $Q$
we  have $$\Lambda_n g :=  2^{-nk} g(x)\sum\limits_{x \in P_n}.1=2^{-nk} g(x)2^{(n-N)k}=2^{-Nk}g(x)$$
$$\implies 2^{-Nk}g(x) \neq \Lambda_N g :=  2^{-Nk} \sum\limits_{x \in P_n} g(x)$$
Therefore $$\Lambda_N g \neq
 \Lambda_n g$$
 A: The issue with your approach is that you’ve factored $g(x)$ out of a sum which is indexed by $x$, but $g$ is not constant on all of $P_n$. However, $g$ is compactly supported and constant on every $Q \in \Omega_N$. This means that:

*

*The sums have finitely many nonzero terms, so you can group terms however you like; and,

*For every $Q \in \Omega_N$, the $2^{(n - N) k}$ points of $P_n$ which lie inside $Q$ all have the same value under $g$, and we can call it $c(Q)$.

Hence, the sum may be regrouped as
\begin{align*}
2^{-nk} \sum_{x\in P_n} g(x) &= 2^{-nk} \sum_{Q\in\Omega_N} \sum_{x\in Q\cap P_n} g(x) \\
&= 2^{-nk} \sum_{Q\in\Omega_N} c(Q) \sum_{x\in Q\cap P_n} 1 \\
&= 2^{-nk} \sum_{Q\in\Omega_N} c(Q) \cdot 2^{(n - N)k} \\
&= 2^{-Nk} \sum_{Q\in\Omega_N} c(Q) \\
&= 2^{-Nk} \sum_{x\in P_N} g(x),
\end{align*}
where the first equality is justified as grouping together all the $x$’s which lie in the same $Q$; and the last equality is justified as each $Q \in \Omega_N$ can be identified by its bottom-left corner $x \in P_N$, and $c(Q) = g(x)$ in that case.
A: Note : $x \in P_N $ not $P_n$, so $P_n$ has exactly $2^{(N-N)k}$ points in $Q$. This implies
$$\Lambda_N g =  2^{-Nk} \sum\limits_{x \in P_N} g(x)
=2^{-Nk} g(x) 2^{(N-N)k}=2^{-Nk}g(x)=\Lambda_n g$$
