# 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!!

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.
• Thank you for the answer, I do appreciate this! I was saying, rather than swapping out uniform convergence, that we can swap out pointwise convergence with in measure convergence, because $X$ is a finite set. Commented Dec 26, 2021 at 22:59