I am working through the proof of Theorem 2.20 in Rudin's Real and Complex Analysis and am getting stuck. The statement of the theorem is as follows: enter image description here

He starts the proof with defining a positive linear functional: enter image description here

He defines $P_n$ earlier in the text as well as some other relevant terms as:

enter image description here

I understand what Rudin is doing with uniform continuity, but I don't understand how property (c) of the collection $\Omega_n$ is used to show that $\land_N g = \land_n g$. Any elaboration on this point would be appreciated.

  • $\begingroup$ $\Lambda$ creates $\Lambda$. $\endgroup$
    – Randall
    Commented Mar 12, 2023 at 0:51

1 Answer 1


According to the property (c), for $n>N$, the set $P_{n}$ has exactly $2^{(n-N)k}$ points in $Q\in\Omega_{N}$. On the other hand, from (a) of 2.19 we know for each $x\in P_{n}$, there is only one $Q\in\Omega_{N}$ such that $x\in Q$. Now $g$ is constant on each $Q$ belonging to $\Omega_{N}$, thus we have$$\begin{aligned}\Lambda_{n}g&=2^{-nk}\sum_{x\in P_{n}}g(x)=2^{-nk}\sum_{Q\in\Omega_{N}}\sum_{x\in P_{n}\cap Q}g(x)\\ &=2^{-nk}\sum_{Q\in\Omega_{N}}2^{(n-N)k}\cdot g|_{Q}=2^{-Nk}\sum_{x\in P_{N}}g(x)\\ &=\Lambda_{N}g.\end{aligned}$$

  • $\begingroup$ Thanks for your answer! I'm still a little confused though. How does the fact that $P_n$ has $2^{(n-N)k}$ points in $Q \epsilon \Omega_N$ give you the middle equality? $\endgroup$ Commented Mar 12, 2023 at 14:40
  • $\begingroup$ @TeaDrinker7 I have edited it. $\endgroup$
    – mio
    Commented Mar 12, 2023 at 16:14
  • $\begingroup$ Okay, I think I follow everything now. Thanks so much for your explanation. $\endgroup$ Commented Mar 12, 2023 at 17:28

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