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!
We can, without loss of generality, assume that the sequence $f_n$ is decreasing and convergent almost everywhere to $0$ (if it's not the case, consider $\displaystyle g_n(x):=\sup_{k\geq n}|f_k(x)-f(x)|$, which is dominated by $2g$, an integrable function. We fix $\varepsilon>0$, and we are looking for a measurable set $A$ such that $\mu(X\setminus A)\leq \varepsilon$ and the sequence $\{f_n\}$ converges uniformly to $f$ on $A$. We have for all integers $j$ and $n$ that $$\mu\left(\left\{g_n\geq \frac 1j\right\}\right)\leq j\int_X |g_n|d\mu,$$ and the monotone convergence theorem gives us that $\displaystyle\lim_{n\to\infty}\int_X |g_n|d\mu=0$. Hence for all $j\geq 1$, we can pick $n_j\in\mathbb N$ such that $$\mu\left(\left\{g_{n_j}\geq \frac 1j\right\}\right)\leq \varepsilon2^{-j}.$$ Now, put $A_j:=\left\{g_{n_j}\geq \frac 1j\right\}$ and $\displaystyle A:=\bigcap_{j\geq 1}\complement_X A_j$. We have $$\mu(X\setminus A)=\mu\left(\bigcup_{j\geq 1}A_j\right)\leq \sum_{j\geq 1}\mu(A_j)=\varepsilon\left(\frac 1{1-2^{-1}}-1\right)=\varepsilon$$ and $\displaystyle\sup_{x\in A}\,|g_{n_j}-0|\leq \frac 1j$. Since the sequence $\displaystyle\left\{\sup_{x\in A}\,g_n(x)\right\}$ is decreasing and has a sub-sequence which converges to $0$, the whole sequence converges to $0$, and we are done.