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Here is theorem 1.14 from Rudin's RCA:

If $f_n : X \to [-\infty, \infty]$ is measurable for $n=1,2,3,...$, and $g = \sup_{n \ge 1} f_n$ and $h = \lim_{n \to \infty} \sup f_n$, then $g$ and $h$ are measurable.

Immediately following this is the corollary (a):

The limit of every pointwise convergent sequence of complex measurable functions is measurable.

I don't see how this is a corollary of theorem 1.14, since the theorem concerns (extended) real-valued functions, whereas the sequence of functions in the corollary take on values in $\Bbb{C}$.

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  • $\begingroup$ Your definition of $h$ makes very little sense, since $\sup f_n$ does not depend on $n$. I know it doesn't say $h=\lim_{n\to\infty}\sup f_n$ in the book - what's the actual definition of $h$? $\endgroup$ – David C. Ullrich Dec 7 '17 at 18:37
  • $\begingroup$ @DavidC.Ullrich I assure you, it is no definition of mine; what I typed is exactly how it appears in the book (at least in my copy). $\endgroup$ – user193319 Dec 7 '17 at 19:02
  • $\begingroup$ I find that very hard to believe. My copy is missing. Maybe it actually says $\displaystyle\limsup_{n\to\infty}$? That's not the same thing as what you wrote - find the definition of "$\limsup$" somewhere... $\endgroup$ – David C. Ullrich Dec 7 '17 at 19:06
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Hint: Apply the theorem for $-f_n$. You will get the inf version and the liminf version. Now when limsup and liminf are measurable, what can you say about the limit when it exists?

Update: Sorry I didn't read your question fully. You wanted to know how this result for reals extends to complex functions. Well, any complex measurable function $f$ is of the form $Re(f) + iIm(f)$. Also if $\lim_n f_n =f$ where $f_n$ are complex measurable functions then note that this is iff $Re(f_{n}) \to Re(f)$ and $Im(f_{n}) \to Im(f)$.

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I think the $h$ should be $h=\lim_{n}\sup_{k\geq n}f_{k}$. If we can show that for each $n$, $\sup_{k\geq n}f_{k}$ is measurable, then the corollary applies to $h$ along with $(\sup_{k\geq n}f_{k})_{n}$.

Now we see that \begin{align*} \left(\sup_{k\geq n}f_{k}\right)^{-1}(a,\infty]=\bigcup_{k\geq n}f_{k}^{-1}(a,\infty]. \end{align*}

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