Dominated Convergence Theorem using Egorov I've been reading on Dominated Convergence Theorem and its proof using Fatou-Lebesgue, but I can't seem to figure out how to do so with Egorov's theorem.
If $\nu$ is a finite Baire measure on a compact Hausdorff space $X$, I first let $U_n$ be a sequence of sets such that $U_{n+1} \subset U_n$ with an additional property that $\nu(\cap_n U_n)=0$ . 
I'm first trying to show that if $f_n$ is a characteristic function of $U_n$, then for any $g\in L^1$, $\lim_n \int | gf_n| \,d\nu=0$ using monotone convergence. 
Any hints would be appreciated. Thank you.
 A: Let $\{f_n\}$ a sequence of integrable functions which converges almost everywhere to $f$, and such that $|f_n(x)|\leq g(x)$, where $g$ is integrable, for almost all $x$ and all $n$. We assume that we work in a finite measured space. Fix $\varepsilon>0$. We can find $\delta>0$ such that if $A$ is measurable and the measure of $A$ is $\leq\delta$ then $\int_Ag(x)d\nu(x)\leq \varepsilon$. Let $C_{\varepsilon}$ a measurable set such that $\nu(X\setminus C_{\varepsilon})\leq\delta$ and $\sup_{x\in C_{\varepsilon}}|f_n(x)-f(x)|\to 0$ (by Egoroff's theorem). We have 
\begin{align}
\left|\int_Xf_n(x)d\nu(x)-\int_Xf(x)d\mu(x)\right|&\leq \nu(C_{\varepsilon})\sup_{x\in C_{\varepsilon}}|f_n(x)-f(x)|+\int_{C_{\varepsilon}}|f_n(x)-f(x)|d\nu(x)
\\
&\leq \nu(X)\sup_{x\in C_{\varepsilon}}|f_n(x)-f(x)|+2\int_{C_{\varepsilon}}|g(x)|d\nu(x)\\
&\leq \nu(X)\sup_{x\in C_{\varepsilon}}|f_n(x)-f(x)|+2\varepsilon,
\end{align}
hence for all $\varepsilon>0$
$$\limsup_{n\to +\infty}\left|\int_Xf_n(x)d\nu(x)-\int_Xf(x)d\mu(x)\right|\leq 2\varepsilon,$$
which gives the conclusion of the dominated convergence theorem. 
A: I think there are several facts which you could apply when proving Lebesgue by Egorov.
1/ Choose a compact set $\Omega$ such that $\int_{R-\Omega} f dx$ is small enough, here $f$ is the dominate function, which is in $L^1$;
2/ for compact $\Omega$, we can apply Egorov theorem, choosing uniform convergence sequence out of a set with small Lebesgue measure;
3/ on the part where the sequence doesn't uniformly converge, the integration is very small, because $\int_{-\infty}^x f(s)ds$ is absolutely continuous.
