Finite a.e. assumption of Egorov I am looking for an example to show that the requirement that $f$ be finite a.e. in Egorov's theorem cannot be dropped. 
I was thinking about $f_n = n$, but here I am not able to see why $f_n$ does not converge to $f$ almost uniformly. (Or rather, I don't know what it means to talk about a sequence of functions converging uniformly to $\infty$).
Any help is appreciated.  

This is exactly the problem - you don't have uniform convergence on any nonempty subset of $\Bbb{R}$.
Note that if a sequence $f_n$ converges uniformly on some nonempty set $A$, then it must satisfy the uniform Cauchy criterion, i.e. $||f_n - f_m||_{L^\infty(A)} \to 0$ as $n,m \to \infty$. However, letting i.e. $m=n+1$, we have $||f_n - f_m||_{L^\infty(A)} = 1$ for any $n$ for any nonempty $A$ (in your example, where $f_n = n$).
