Proof of measurability of the sum of extended-real measurable functions I was reading Rudin's Real and Complex Analysis (3rd ed.), and on page 19 he says: (The key point here is about arithmetic in $[0, \infty]$.)

Observe that the following useful proposition holds: (with arithmetic in $[0, \infty]$)
If $0\le a_1 \le a_2 \le \cdots$, $0\le b_1 \le b_2 \le \cdots$, $a_n \to a$ and $b_n \to b$, then $a_nb_n\to ab$.
If we combine this with Theorems $1.17$ and $1.14$, we see that sums and products of measurable functions into $[0, \infty]$ are measurable.

Theorem 1.14 If $f_n:X\to [-\infty, \infty]$ is measurable, and $g=\sup_n f_n$, then g is measurable.
Theorem 1.17 Let $f:X\to [0, \infty]$ be measurable.  There exist simple measurable funcctions $s_n$ on $X$ s.t. $0\le s_1\le s_2\le \cdots\le f$, and $\forall x\in X: s_n(x)\to f(x)$ as $n\to \infty.$

I am not very sure what he meant.  I list my attempted proof below, and would appreciate it if someone can confirm if it's valid, or point out if I miss his point.

Suppose $f, g:X\to[0, \infty]$ are two measurable functions.  I want to show that $f+g$ is also measurable.  I'd proceed as follows:
First, I state an easily proved proposition similar to the one given above: If $0\le a_1 \le a_2 \le \cdots$, $0\le b_1 \le b_2 \le \cdots$, $a_n \to a$ and $b_n \to b$, then $a_n+b_n\to a+b$.  (Even if $a$ or $b$ takes value of $\infty$.)
Since $f, g$ are measurable, by Theorem 1.17, there must exist simple measurable functions $f_n, g_n$ on $X$ s.t. $0\le f_1 \le f_2 \le \cdots\le f$, and $\forall x\in X:f_n(x)\to f(x).$  Similarly, $0\le g_1 \le g_2 \le \cdots\le g$, and $\forall x\in X: g_n(x)\to g(x).$  (In Rudin's book, simple functions are real functions, not extended-real.) Hence $\forall x\in X:f_n(x)+g_n(x)\to f(x)+g(x)$ by the proposition above.  (Note that $f_n+g_n$ is real, while $f+g$ is extended-real valued.) But $f_n, g_n$ are simple measurable functions, so their sum $f_n+g_n$ must be simple and measurable.  Moreover, $\{f_n+g_n\}$ is a sequence of increasing functions, so $$\lim_{n\to \infty} (f_n+g_n)=\sup_n(f_n+g_n).$$
Since the supremum of a sequence of real measurable functions must be measurable (Theorem 1.14), we conclude that $f+g$ is measurable (even though it's extended-real).
 A: You are correct!

I have just one comment. Perhaps it will be better to say that
$$
\lim_{n \to \infty} (f_n + g_n) = \limsup_{n \to \infty} (f_n + g_n) \tag{1}
$$ 
instead of
$$
\lim_{n \to \infty} (f_n + g_n) = \sup_n (f_n + g_n). \tag{2}
$$
This is because for every convergent sequence $\{a_n\}$,
$$
\lim_{n \to \infty} a_n = \limsup_{n \to \infty} a_n,
$$
whereas only if $\{a_n\}$ is a monotonically increasing sequence that is convergent is it true that
 $$
\lim_{n \to \infty} a_n = \sup_{n \geq 1} a_n.
$$
So, using (2) requires a tiny bit more of an explanation than (1).
The measurability of $\limsup_{n \to \infty} (f_n + g_n)$ also follows from Theorem 1.14 (you haven't stated it in full in the question details):

Theorem 1.14. If $f_n : X \to [-\infty,\infty]$ is mesaurable, for $n = 1,2,3,\dots$, and $$g = \sup_{n \geq 1} f_n, \qquad h = \limsup_{n \to \infty} f_n,$$ then $g$ and $h$ are measurable.


For completeness, here's an example of a sequence $\{ a_n \}$ such that $\sup_{n \geq 1} a_n \neq \limsup_{n \to \infty} a_n$: take the sequence given by $a_n = 1/n$. Then the supremum is $1$ whereas the upper limit is $0$.
