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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 $}.

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  • $\begingroup$ Are you asking for verification of your proof? $\endgroup$ – walkar May 5 '16 at 1:30
  • $\begingroup$ Yes .Please do tell if you find a flaw. $\endgroup$ – ibnAbu May 5 '16 at 1:37
  • $\begingroup$ I haven't seen a proof with outer measures, so I'm not sure. $\endgroup$ – walkar May 5 '16 at 1:41
  • $\begingroup$ @walkar. $B_n$ is measurable so its outer and inner measures are equal $\endgroup$ – ibnAbu May 5 '16 at 1:49
  • $\begingroup$ $A_n$ doesn't appear to be well defined. You have it set to an infinitude of (possibly distinct) sets for each choice of $m \geq n$. $\endgroup$ – Eric Towers May 5 '16 at 1:53
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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 $\int_0^1 0\, dx = 0.$

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  • $\begingroup$ Not a counter example.Your $ f_n $ is not bounded by any real number nor is your {$f_n$} uniformly integrable. $\endgroup$ – ibnAbu May 5 '16 at 11:05
  • $\begingroup$ 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. $\endgroup$ – zhw. May 5 '16 at 19:23
  • $\begingroup$ I gave a proof of the theorem used in the proof : math.stackexchange.com/questions/1772057/… $\endgroup$ – ibnAbu May 7 '16 at 18:38
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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 $}.

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

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