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Let $(f_n)_{n \in \Bbb N}$ be a sequence of measurable functions on $E$, that converges almost everywhere pointwise towards $f$.
Let $(g_n)_{n \in \Bbb N}$ be a sequence of integrable functions on $E$ that converge almost everywhere on $E$ pointwise towards $g$.
Also, suppose that $|f_n| \le g_n$ $\forall n \in \Bbb N$.
I have to show that:$$\lim_{n\to \infty} {\int_E g_n } = \int_E g \Rightarrow \lim_{n\to \infty} {\int_E f_n } = \int_E f$$

I don't really understand how i should show this. I don't see why the right hand side of the relation isn't just allways true due to dominated convergence (i can simply pick one $g_n$). Any ideas or tipps on how to show this and on why it isn't already true because of dominated convergence?

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  • $\begingroup$ You can't simply pick one $g_n$, because for $m\neq n$ you don't know if $|f_m|<g_n$ $\endgroup$
    – user160738
    Dec 11, 2016 at 9:02
  • $\begingroup$ Indeed this is a stronger result than dominated convergence. Luckily, you can just copy the proof of dominated converge with some obvious modifications to show this result! $\endgroup$
    – air
    Dec 11, 2016 at 9:04
  • $\begingroup$ I don't understand why to say that $g_{n}$-s converges almost everywhere pointwise. Why just not to say that is converges almost everywhere(without using "pointwise", as it is obvious). Yes you can take on g, $h(x)=sup_{n}{g_{n}(x)}$. What I mean, is that $ |f_{n}(x)| \leq h(x) $ which is convergent almost everywhere as well. $\endgroup$
    – kolobokish
    Dec 11, 2016 at 9:13

1 Answer 1

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For the Dominated Convergence to apply, you must have that the $f_n$'s are dominated by a function $\varphi$ whose integral is finite.

Now I added one extra condition to the given because I don't see how it is possible without it. So all in all the conditions become

  • $\lim_{n\to\infty}f_n=f$
  • $\lim_{n\to\infty}g_n=g$
  • $|f_n|\leq g_n$.
  • $\lim_{n\to\infty}\int g_nd\mu=\int g d\mu$
  • $\int g d\mu<\infty$ (extra condition)

Now notice that it is enough to show
$$\lim_{n\to\infty}\int|f_n-f|d\mu=0 $$ The result you want follows immediately from the triangle inequality. To start the proof, define $$\varphi_n=g_n+g-|f_n-f| .$$ Then $\varphi_n$ is positive measurable since
$$|f_n|\leq g_n\implies|f|\leq g\implies\varphi_n\geq g_n+g-(|f_n|+|f|)\geq g_n+g-(g_n+g)=0.$$ Also notice that $\varphi_n\to 2g$ as $n\to\infty$ almost everywhere on $E$. Therefore, we can use Fatou's Lemma to deduce $$\int 2g d\mu=\int2\lim_{n\to\infty}g_nd\mu= \int\lim_{n\to\infty}\varphi_nd\mu= \int\liminf_{n\to\infty}\varphi_nd\mu\leq\liminf_{n\to\infty}\int \varphi_nd\mu,$$ and furthermore, $$\varphi_n\leq g_n+g\implies\int \varphi_nd\mu\leq\int(g_n+g)d\mu\implies \limsup_{n\to\infty}\int \varphi_nd\mu\leq\int 2gd\mu.$$ Hence $$\lim_{n\to\infty}\int \varphi_nd\mu=\int 2g d\mu.$$ Since $$\int (g_n+g) d\mu=\int\varphi_nd\mu+\int |f_n-f|d\mu,$$ then by letting $n\to\infty$ we see that $$\int 2gd\mu=\int 2gd\mu+\lim_{n\to\infty}\int|f_n-f|d\mu.$$ Before you rush and do the cancellation, you need to make sure that $\int2gd\mu$ is finite which is why I needed the extra condition. You are done. Notice that this is very similar to the proof of the Dominated Convergence.

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    $\begingroup$ And without the condition $\int g\,d\mu < +\infty$, the conclusion need not hold, let $f_n = \chi_{[n,n+1]}$ and $g_n = \chi_{[0,2^n]}$ for an example. $\endgroup$ Dec 11, 2016 at 11:30
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    $\begingroup$ You can shorten the last part using $$\liminf_{n\to\infty} \int \varphi_n\,d\mu = 2\int g\,d\mu - \limsup_{n\to\infty} \int \lvert f_n - f\rvert\,d\mu.$$ $\endgroup$ Dec 11, 2016 at 11:34

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