# 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 $$\Lambda_N g \neq \Lambda_n g$$

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:

1. The sums have finitely many nonzero terms, so you can group terms however you like; and,
2. 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.

• thanks u @Shoteyes Can you recommend a good book for measure theory ? Im weak in measure theory Jun 2, 2021 at 17:06
• Honestly, I like Rudin’s RCA even though it’s not strictly a measure theory book, since it covers so much of real and complex analysis. The only qualm I have with it is that some things lack motivation, but drawing pictures yourself can help with that. Jun 2, 2021 at 17:10

Note : $$x \in P_N$$ not $$P_n$$, so $$P_n$$ has exactly $$2^{(N-N)k}$$ points in $$Q$$. This implies

$$\Lambda_N g = 2^{-Nk} \sum\limits_{x \in P_N} g(x) =2^{-Nk} g(x) 2^{(N-N)k}=2^{-Nk}g(x)=\Lambda_n g$$