Different versions of Egoroff's theorem

Question

I have seen many versions of Egoroff's theorem which do not require that $f_k$ and $f$ to be finite a.e.. But in most of the proof, the first steps are setting $A_{i,j}=\cup_{k\geq j}\{x;|f_k(x)-f(x)| \geq 1/2^i\}$. And the set is not well-defined when $f_k$ and $f$ are infinite. Is there any proof concerning this case?

Answer 1

The general proof is like this: When we say $f_{n}\rightarrow f$ pointwise a.e., we mean under the real numbers convergence.

Let $A=\{x\in X:\lim_{n\rightarrow\infty}f_{n}(x)=f(x)\}$, so for each $x\in A$, $f(x)$ is a real number, so there is some $n_{x}$ such that $f_{n}(x)$ is a real number for all $n\geq n_{x}$.

Now $\mu(X-A)=0$ by assumption and we let \begin{align*} A_{k,m}=\bigcap_{n\geq m}\left\{x\in A:|f_{n}(x)-f(x)|<\dfrac{1}{k}\right\}, \end{align*} this set $A_{k,m}$ is well defined, because it is a subset of $A$, and $f(x)$ is a real number for $x\in A$, so $|f_{n}(x)-f(x)|$ is defined as an extended real number, in other words, it is not of the form $\infty-\infty$.

Now we can prove that $\lim_{m\rightarrow\infty}\mu(X-A_{k,m})=0$ for each fixed $k$ and we are ready to derive the uniform convergence on some measurable set.