# Show that for $|f_n| \le g_n$ $\forall n$: $\lim_{n\to \infty} {\int_E g_n } = \int_E g \Rightarrow \lim_{n\to \infty} {\int_E f_n } = \int_E f$

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?

• You can't simply pick one $g_n$, because for $m\neq n$ you don't know if $|f_m|<g_n$ Dec 11, 2016 at 9:02
• 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!
– air
Dec 11, 2016 at 9:04
• 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. Dec 11, 2016 at 9:13

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.

• 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. Dec 11, 2016 at 11:30
• You can shorten the last part using $$\liminf_{n\to\infty} \int \varphi_n\,d\mu = 2\int g\,d\mu - \limsup_{n\to\infty} \int \lvert f_n - f\rvert\,d\mu.$$ Dec 11, 2016 at 11:34