TITLE: Egorov's Theorem - Counterexample in Infinite Case
QUESTION [10 upvotes]: Why does Egorov's theorem not hold in the case of infinite measure?  It turns out that, for example, $f_n = \chi_{[n,n+1]}x$ does not converge nearly uniformly, that is, it does not converge on E such that for a set F m(E\F) < $\epsilon$.  Is this simply true because it takes on the value 1 for each n but suddenly hits 0 when n ---> infinity?

REPLY [16 votes]: Set $\forall n\geq1: f_n:[0,\infty[\to\{0,1\}, f_n:=\chi_{[n-1,n]}$. Then $f_n\to0$ pointwise on $\mathbb{R}$. Suppose $\exists F\subseteq\mathbb{R}: f_n\stackrel{u.}{\to}0$ on $F$, i.e. that
$$\forall \epsilon>0,\exists N,\forall n\geq N,\forall x\in F: |f_n(x)|<\epsilon.$$
For $\epsilon:=1, \exists N,\forall n\geq N,\forall x\in F:|f_n(x)|<1$, so that $x\not\in[N,\infty[$. Thus $F\subseteq [0,N[$, and consequently $m(\mathbb{R}-F)\geq m([N,\infty[)=\infty$.

The moral of the story is then that for this example any set on which we have uniform convergence has to be of finite measure, and we cannot make the rest have arbitrarily small measure.