Let (X,F,m) be a measure space such that m(X) < oo. Let fn be a
sequence of measurable functions which converges pointwise a.e. to a
real valued function f on X. Show that for every e > 0 there is a
measurable set E C X such that m(X \ E)<e and such that (fn) converges
uniformly to f on E. (This result is known as Egorov's theorem) [Hint:
You may assume that fn ? > f everywhere on X. (Why?) For each positive
integer n and k, consider the sets
 
E(n,k)= UNION from m=n to oo{xEX: |fm(x) - fn(x}| >= 1/k}
 
The sets E(n, k) decrease as n increases, and INTERSECT N=1 to oo
E(n,k)= (Why?) Thus m(E(n, k)) ?>0 as n ?> oo. (Why?) Choose a
subsequence (nk) such that m(E(nk,k) < e/2^k. (How?) Let E = UNION
from k=1 to oo E(nk,k). Show that m(E} < e. Now, if x is not in E,
then x is not in E(nk, k) for any k, and so |fm(x) ~ f(x)| < l/k for
all m >= nk]  .... (oo is infinite, nk is n sub k, e is epsilon)