Egorov's Theorem - Counterexample in Infinite Case

Solution 1:

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.

Solution 2:

$f_n$ converges pointwise to the zero function on $\mathbb{R}$ (here $E = \mathbb{R})$. However there doesn't exist a set of finite measure $F$ such that $f_n$ converges uniformly on $\mathbb{R} \setminus F$. To see this note that for large enough $n$, $f_n$ will take both the values $0$ and $1$ on $\mathbb{R} \setminus F$.