Why do we need a finite measure to have this theorem?

Question

Here is the theoerem 6.2.4 given in “Ross Leadbetter -A basic course of measure and probability”

Let $(X,\mathfrak{M},\mu)$ be a finite measure space.

Let $f_n,f$ be measurable (real) functions on $X$.

Then, $f_n\to f$ $\mu$-a.e. iff $\mu(\limsup_n \{x:|f_n(x)-f(x)|\geq \epsilon\})=0$ for each $\epsilon>0$.

This seems obviously true for arbitrary measure space. However, the author emphasize that this thereom holds for a finite measure space.. Why do we need a finiteness condition here? Am I missing something?

Answer 1

I think your use of the word "obviously" is not warranted, but you are right, the equivalence holds whether or not $\mu$ is finite.

If we define

$$N = \bigl\{ x : \bigl(f_n(x)\bigr) \text{ does not converge to } f(x)\bigr\}$$

and

$$M(\epsilon) = \bigl\{ x : \lvert f_n(x) - f(x)\rvert \geqslant \epsilon \text{ for infinitely many } n\bigr\}\,,$$

then note that $\epsilon_1 \leqslant \epsilon_2 \implies M(\epsilon_2) \subset M(\epsilon_1)$ implies

$$N = \bigcup_{\epsilon > 0} M(\epsilon) = \bigcup_{k = 0}^{\infty} M(2^{-k})\,,$$

it is clear that the equivalence

$$\mu(N) = 0 \iff \mu\bigl(M(\epsilon)\bigr) = 0 \text{ for all } \epsilon > 0$$

holds.