Shortest Proof of Lebesgue Dominated Convergence Theroem ( 5 lines) without using Fatou's lemma

Question

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 M$

The condition that each $ G_n $ is bounded by M can be relaxed to uniform integrability of {$ G_n $}.

Answer 1

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 M$

The condition that each $ G_n $ is bounded by M can be relaxed to uniform integrability of {$ G_n $}.

Answer 2

Counterexample: On $(0,1),$ let $f_n(x) = n^2x^n.$ Then each $f_n$ is bounded on $(0,1),$ and $f_n \to 0$ pointwise on $(0,1).$ But $\int_0^1 f_n(x)\, dx = n^2/(n+1) \to \infty,$ while $\int_0^1 0\, dx = 0.$