Fine Print of Egorov's Theorem

Question

The Egorov's theorem in my textbook requires that the function to be define on a set with finite measure. Why is this necessary, please? Thank you!

Answer 1

Take e.g. $f_n = \chi_{[n,n+1]}$. Then $f_n \rightarrow 0$ pointwise, but there is no set $E$ of (Lebesgue) measure $|E| < 1/2$, so that $f_n \rightarrow 0$ uniformly on $\Bbb{R} \setminus E$ (why?).

EDIT: If you look at the proof (e.g. http://en.wikipedia.org/wiki/Egorov%27s_theorem#Proof ), then "continuity of $\mu$ from above" is used, e.g. that $\mu(\bigcap_n E_{n,k}) = \lim_n \mu(E_{n,k})$ (here we also use $E_{n+1,k} \subset E_{n,k}$ for all $n,k$).

But this is in general not true if the measure space is of infinite measure, e.g. $|\bigcap_n [n,\infty)| = 0$, but $|[n, \infty)| = \infty$ for all $n$.