Why is this a contradiction? Egoroff's Thm when $m(E) = \infty$ Problem 27: Show that the conclusion of Egoroff’s Theorem can fail if we drop the assumption that the domain has finite measure.
Solution: Consider sequence $f_n(x) = χ_{[n,\infty)}(x)$. Clearly $f_n \to 0$ pointwise on $\mathbb{R}$. Suppose that there existed a set $F$ such that $m(\mathbb{R} \setminus F) < \epsilon$ and $f_n \to 0$ uniformly on $F$. Since $f_n$ is an indicator function, this means we can find an $N$ such that $f_N = 0$ on $F$. This implies $F \subset (-\infty, N)$ and so $[N, \infty) \subset \mathbb{R} \setminus F$, giving $m([N, \infty)) \le m(\mathbb{R} \setminus F) < \epsilon$, which is a contradiction.
I don't understand the last line.  What is being contradicted?  Does $f$ not converge uniformly?  Why? 

It would have been clearer to write 
$$\infty = m([N,\infty)) \leq m(\mathbb{R} \setminus F)<\epsilon.$$
The end result of this little calculation is that if $F$ is any set on which $f_n \to 0$ uniformly then $m(\mathbb{R} \setminus F)=\infty$. 
It also would have been slightly better use of logic to fix some $\epsilon$ (since here $\epsilon$ is actually a totally unbound variable), but that's kind of nitpicking.
