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

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 {$G_n$}.

• Are you asking for verification of your proof? – walkar May 5 '16 at 1:30
• Yes .Please do tell if you find a flaw. – ibnAbu May 5 '16 at 1:37
• I haven't seen a proof with outer measures, so I'm not sure. – walkar May 5 '16 at 1:41
• @walkar. $B_n$ is measurable so its outer and inner measures are equal – ibnAbu May 5 '16 at 1:49
• $A_n$ doesn't appear to be well defined. You have it set to an infinitude of (possibly distinct) sets for each choice of $m \geq n$. – Eric Towers May 5 '16 at 1:53

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 $\int_0^1 0\, dx = 0.$

• Not a counter example.Your $f_n$ is not bounded by any real number nor is your {$f_n$} uniformly integrable. – ibnAbu May 5 '16 at 11:05
• I addressed the result you first claimed, which was in error. So a better comment from you, I think, would have been to admit the error. – zhw. May 5 '16 at 19:23
• I gave a proof of the theorem used in the proof : math.stackexchange.com/questions/1772057/… – ibnAbu May 7 '16 at 18:38

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 {$G_n$}.

• Your proof is false in many ways. – Michael Mar 24 '17 at 14:04