How does countable additivity imply convergence in Rudin's theorem 11.3?

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In Principles of Mathematical Analysis, Walter Rudin has the following theorem 11.3 on page 302 in his book:

Rudin theorem 11.3

I am having great difficulty understanding this theorem. I'll take you through my logic and can someone please tell me where I went wrong.

  • Assuming that if $A_1 \subseteq A_2$ then the union of $A_1$ and $A_2$ is $A_2$, then it seems to me that $A = A_{\infty}$.

  • If that is true, then it seems to me like $A_{\infty}$ = $\bigcup_{i=1}^\infty A_i $

  • And if that is true, then we seem to just be saying $\phi(A_{\infty})$ = $\phi(\bigcup_{i=1}^\infty A_i)=\phi(A)$.

In which case, Why do we even need to take the limit? The result of the theorem is achieved just by plugging in different equalities for $A$.

Where did I go wrong in my logic?

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There is no "$A_\infty$". When the theorem statement says "$n=1,2,3,\dots$", this means that $n$ ranges over all positive integers, so $\infty$ is not a valid value of $n$ and there is no set "$A_\infty$" given in the statement.

In any case, I have no idea how you think that the theorem would follow by "plugging in different equalities" instead of taking a limit. The very statement of the theorem is about a limit: it claims that $\phi(A_n)$ converges to $\phi(A)$ as $n\to\infty$.