Lebesgue Dominated Convergence Theorem other proof I have read lots of proofs for the Lebesgue Dominated Convergence Theorem and a lot of them involves Fatou's Lemma. My question is can we directly prove LDCT without using Fatou's Lemma? We study measure theory but we did not cover Fatou's Lemma yet and I dont have any idea how to start the proof. 
My first guess is we use Monotone Convergence Theorem but the monotonicity of the sequence is not part of the assumption. I don't know any other theorem that we discussed that involves limits hence I do not know where to start. 
Any lead, hint or help will be much appreciated. Thank you!
 A: There's  a very simple proof of DCT for sums, where you start by choosing $N$ with $\sum_{n>N}g(n)<\epsilon$. You can generalize this to any measure space using Egoroff's  theorem:
Say $g\ge0$, $|f_n|\le g$ and $f_n\to f$ almost everywhere. Since $f_n=0$ on the set where $g=0$ we can ignore that set and assume, just to simplify the notation, that $g>0$ everywhere. Let $\epsilon>0$.
There exists $\delta>0$ so that $$\int_Eg<\epsilon$$for any $E$ with $\mu(E)<\delta$ (and btw this is easy to prove using nothing but the fact that you can approximate $g$ is $L^1$ by simple functions, more or less by definition.) Now Egoroff, which again is quite elementary, shows that there exists $E$ with $\mu(E)<\delta$ such that $f_n/g\to f/g$ uniformly on $X\setminus E$. So for large enough $n$ we have
$$\int|f_n-f|\le 2\int_E g+\int_{X\setminus E}\left|\frac{f_n}g-\frac fg\right|g\le 2\epsilon+\epsilon||g||_1.$$
(In fact you don't really need Egoroff: You could just let $E_n$ be the set where $|f_n-f|/g>\epsilon$ and note that $\mu(E_n)\to0$.)
A: It is possible to prove Dominated Convergence Theorem without dwelling much on the machinery of Measure Theory.
The proof relies on see this theorem
If  {$G_n$} is a sequence of bounded measurable functions  and $ | G_n | \le M $ where M is a positive real number $\lim\limits_{n\mapsto \infty} G_n =F$ on a bounded measurable set E , $\epsilon> 0 $
Let $A_n =$ { $x : |G_m(x)-F(x)|<\epsilon$} whenever  $m \ge n$
and let $B_n =$ { $x : |G_n(x)-F(x)|<\epsilon$} 
Then $A_n\subseteq A_{n+1}\subseteq B_{n+1}\subseteq E   , \lim\limits_{n\mapsto \infty} \mu^*( A_n) = \mu^*( E) $  according to see here
Then$\lim\limits_{n\mapsto \infty} \mu^*( B_n) =\mu^*( E)  , \lim\limits_{n\mapsto \infty} \mu^*( B_n^C) =0  $
$B_n$ is measurable so its outer and inner measures are equal
$ \lim\limits_{n\mapsto \infty}  | \int_{E}(G_n -F) | \le\lim\limits_{n\mapsto \infty}\int_{B_n}|(G_n -F)|+\lim\limits_{n\mapsto \infty}\int_{B_n^C}|(G_n -F)| \le \epsilon \mu (E)$
The condition that each $ G_n $ is bounded by $M$ can be relaxed to uniform integrability of {$ G_n $}.  The sequence {$ G_n $} if  bounded by some integrable function is a special case of uniform integrability.
