Question on Egorov's Theorem: How do you find such $E_{\epsilon}$ sets?

Egorov's Theorem is as follows: Let $X$ be a finite measure space. If $f_n\rightarrow f$ pointwise a.e., then for all $\epsilon>0$, there exists $E_{\epsilon}\subset X$ such that $m(E_{\epsilon})<\epsilon$ and $f_n\rightarrow f$ uniformly on $X\setminus E_\epsilon$.

I don't have a specific problem in front of me, but suppose we had a problem where we were given some hypothesis, and we had to find such a $E_\epsilon$. Is there a standard method to find such a set? Could you maybe provide some examples? I (believe I) understand the proof of Egorov's theorem, but I am having a hard time actually finding such sets, rather than just claiming such sets exist. One last question: since $X$ is a finite measure space, then we can use $f_n\rightarrow f$ in measure in the original claim, correct? Thank you!!

Solution 1:

Nope, no "general" method to find these sets, Egorov's theorem just proves that they exist. For a loose strategy, if you have some functions $(f_n)_n$ which converge pointwise but not absolutely, typically they converge quite fast everywhere except a "small set". Finding a small set where $f_n$ takes "abnormal" values and removing it will make the $f_n$s converge uniformly. As an example, Consider the measure space $\big((0,1),\mathscr{L}_{(0,1)},\lambda\big)$ and the sequence of measurable functions $f_n(x)=x^n$. Then, $f_n\to 0$ pointwise but not uniformly. However, if we consider the Lesbesgue measurable set $(0,1-\epsilon)$ for some $\epsilon\in(0,1)$ then $f_n\to 0$ uniformly on this set. How did I choose this set? Well, if you look at the graph of say, $x^{20}$,

Then you can see that it is really small for everything to the left of, say $0.8$, but it increases rapidly, i.e, is far away from zero, for the right little bit. So in this case you choose to remove the bit of the graph where $f_n$ is "far away" from zero.

Finally, something I'd like to clarify. Even on a finite measure space, $f_n\to f$ in measure does NOT, in general, imply uniform convergence. As a counterexample consider the sequence of simple functions $f_n=\mathbf{1}_{(0,1/n)}$. Then $f_n\to 0$ in measure, but not uniformly (check this!). So in your statement of Egorov's theorem, NO, you cannot swap out uniform convergence for convergence in measure, because they are not the same thing.