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).


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 $0$.

  • $\begingroup$ For an increasing and bounded sequence, it's well-known that its limit is the supremum, e.g. en.wikipedia.org/wiki/Monotone_convergence_theorem. If the sequence is unbounded and we consider the arithmetic in $[0, \infty]$ as in Rudin's book, it still holds, doesn't it? $\endgroup$ – syeh_106 Jul 27 '18 at 2:20
  • $\begingroup$ @syeh_106 ah, yes of course. Let me modify my answer. Thanks! $\endgroup$ – Brahadeesh Jul 27 '18 at 2:47

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