Omitting the hypotheses of finiteness of the measure in Egorov theorem I want to prove that if I omit the fact that $\mu (X) < \infty$ in Egorov theorem and place instead that our functions $|f_n| <g$ and $g$ is integrable, we still get the result of Egorov's theorem.
Fix $m$ a natural number.
I took $ w_{n} = |f_n-f|$ and thus by DCT $\int |f_{n} - f|$ goes to zero.
Then I took $\bigcup_n {( w_{n} \geq 1/m)}$. I need its measure to be finite. Its measure is less than the sum of the measures of each $ w_n\geq 1/m)$ varying $n$, and by Tchebychev, this is less than $m\int|f_{n} - f|$.  But I got stuck here.
Any help is appreciated. Thanks!

Let $\epsilon>0$ be given.
Define the sets $G_m := \{x: g(x)>1/m\}$. Since $\int g <\infty$, we have $\mu(G_m)<\infty$ for all $m$. For each $m$, apply Egoroff's Theorem to find a $E_m\subseteq G_m$ such that $\mu(E_m)=\epsilon/2^m$ and $f_n \to f$ uniformly on $G_m-E_m$.
Claim: Let $E=\cup_{m=1}^{\infty}E_m$ and we note that $\mu(E)\le \epsilon$. I claim that $f_n\to f$ uniformly on $X-E$.
Indeed, it suffices to show that for every given $k$, there exists $N$ such that $|f_n-f| < 2/k$ on $X-E$ for all $n\ge N$. 


*

*On $X-G_k$, we have $|f_n|,|f|\le g \le 1/k$ so $|f_n-f| < 2/k$ holds for every $n$. 

*On $G_k - E$, which is a subset of $G_k - E_k$, we have that $f_n \to f$ uniformly, so we simply pick $N$ sufficiently large.

