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

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2 Answers 2

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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.

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  • $\begingroup$ thanks u @Shoteyes Can you recommend a good book for measure theory ? Im weak in measure theory $\endgroup$
    – jasmine
    Jun 2, 2021 at 17:06
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    $\begingroup$ 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. $\endgroup$
    – shoteyes
    Jun 2, 2021 at 17:10
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Note : $x \in P_N $ not $P_n$, so $P_n$ has exactly $2^{(N-N)k}$ points in $Q$. This implies

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

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