I was reading the proof of Egorov's theorem and ran into two unclear moments.
1) Why the sets $E_k^n$ are measurable? I was trying in this way: Since $f_j(x)$ are measurable then limit function $f(x)$ is also measurable. Hence $f_j(x)$ and $-f(x)$ are measurable functions. In order $f_j(x)-f(x)$ to be measurable they should be finite a.e. But this condition does not follow the theorem's condition.
2) We know that when we fix $n$ we have $E_k^n\to E$ as $k\to\infty$. Then we can find $k_1$ such that $$m(E-E_{k_1}^n)<\frac{1}{2},$$ and taking $n=1$ we get what we need, namely $m(E-E_{k_1}^1)<\frac{1}{2},$ and we can do this repeatedly and get needed sequence $k_n$. Am I right?
Would be very grateful for help!
1) I don't understand your issue here. The sum/difference of two measurable functions is measurable. Also, if the difference wasn't finite a.e., that would imply at least one of $f_j(x)$ and $f(x)$ are not finite almost everywhere (the union of two measure 0 sets is measure 0 after all).
2) Presumably yes, although it would help to explain what "Corollary 3.3" is.