The problem arises from Walter Rudin's Principles of Mathematical Analysis:
My Background: Rudin's PMA Chapter 1-7.
My Question: How does "(103) almost everywhere on X" follow from (102)?
My Attempt: By Theorem 11.30, we may interchange the summation and integration in (102): $\int_X\sum_{k=1}^{\infty}|g(f_{n_k}-f_{n_k+1})|d\mu\leq\lVert g\rVert<+\infty$, which "seems" to imply $\sum_{k=1}^{\infty}|g(f_{n_k}-f_{n_k+1})|<+\infty$ almost everywhere on $X$.
Claim: $\sum_{k=1}^{\infty}|g(f_{n_k}-f_{n_k+1})|<+\infty$ almost everywhere on $X$. $(*)$
Proof: Suppose, for the sake of contradiction, that for all measurable $A\subset X$, either $\mu(A)\neq0$ or $\sum_{k=1}^{\infty}|g(f_{n_k}-f_{n_k+1})|=+\infty$ for some $x\in X-A$. Note $\mu(\emptyset)=0$, thus there is some $x\in X-\emptyset=X$ such that $\sum_{k=1}^{\infty}|g(f_{n_k}-f_{n_k+1})|=+\infty$. Let $E=\{x\in X|\sum_{k=1}^{\infty}|g(f_{n_k}-f_{n_k+1})|=+\infty\}$. I wish to show that $\mu(E)\neq0$, because then we'd have $\int_E \sum_{k=1}^{\infty}|g(f_{n_k}-f_{n_k+1})|d\mu=+\infty$, which would imply that $\int_X \sum_{k=1}^{\infty}|g(f_{n_k}-f_{n_k+1})|d\mu=+\infty$. But how? I'm not even sure that if $E$ is measurable. Screeching halt.
Then I realized that even IF I proved $(*)$, it still does not lead us to "$|g|\sum_{k=1}^{\infty}|f_{n_k}-f_{n_k+1}|<+{\infty}$ almost everywhere".
Any hint would be greatly appreciated.