# Question on Proof of Theorem 2.20 in Walter Rudin's Real and Complex Analysis

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:

He starts the proof with defining a positive linear functional:

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

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.

• $\Lambda$ creates $\Lambda$. Commented Mar 12, 2023 at 0:51

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}
• 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? Commented Mar 12, 2023 at 14:40