In Rudin's RCA, he defines a function $f$ defined on a set $E\in M$ ($M$ is a $\sigma$-algebra of measurable sets on a measure space $X$) as measurable on $X$ if $\mu(E^c)=0$ and if $f^{-1}(V)\cap E$ is measurable for all open sets $V$ on $E$ and $\mu$ is a measure on $X$.
But then later, he poses a theorem:
Suppose {$f_n$} is a sequence of complex measurable functions defined a.e. on $X$ such that
$$\sum^{\infty}_{n = 1}=\int_X|f_n(x)|d\mu\lt\infty.$$ Then the series $f(x) = \sum^{\infty}_{n=1}f_n(x)$ converges for almost all $x$, $f \in L^1(\mu)$, and
$$\int_Xfd\mu=\sum^{\infty}_{n=1}\int_{X}f_nd\mu.$$
In the proof of this theorem, he states "Let $S_n$ be the set on which $f_n$ is defined, so that $\mu(S^c_n) = 0$ Put $\phi(x)=\sum|f_n(x)|$ for $x\in S=\bigcap S_n$. Then $\mu(S^c) = 0$ and $\int_S\phi d \mu\lt \infty.$ If $E = \{x\in S :\phi (x) \lt \infty\}$, it follows that $\mu(E^c) = 0$."
Why does the space on which $\mu=0$ keep changing in this proof? When he says "Let $S_n$ be the set on which $f_n$ is defined, so that $\mu(S^c_n)=0$," I thought $S_n$ was a fixed but arbitrary function that $f_n$ was defined on, but then $S_n$ may contain something that $S=\bigcap S_n$ does not, so how did he determine that $\mu(S^c) = 0$? This would make sense if the measure function $\mu$ was in reference to $f_n$ the first time but in reference to $\phi$ the second time or if it was a different measure function. I noticed he does the same thing again when he defines the set $E$ as a subset of $S$, so the space for which $\mu$ does not equal $0$ becomes even smaller. Can someone please explain this point to me or clarify how it is possible that $\mu(S_n^c), \mu(S^c),$ and $\mu(E^c)$ all equal $0$?