# Excercise, measure theory

I need help with this excercise:

This excercise is in a chapter where I learn the monotone convergence theorem and Fatous lemma, so I assume I shall use them.

Since $\textbf{1}_Ef_n\rightarrow \textbf{1}_Ef$, this excercise would have been very easy if the sequences were increasing, sadly they are not.

Using Fatou's lemma I get:

$\liminf_{n \rightarrow \infty} \int_Ef_nd\mu \ge \int\liminf_{n \rightarrow \infty}\textbf{1}_Ef_n d\mu=\int_Ef d\mu$.

If I could get another with lim sup, I would maybe be able to be done.

Another problem is that since the $f_n$'s may not be increasing, I do not really know that:

$lim_{n \rightarrow \infty}\int_Ef_nd\mu$ even exist?, because this sequence may not even be monotone?

I guess I should use the part where the entire integral is less than infinity. But I do not really see how to use it. I guess this tells me that the function is finite a.e., but how does that help?

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You could consider the set of all $x$ where there is a subsequence of $f_n(x)$ that is increasing and its complement (where you know that you have a decreasing subsequence). – xavierm02 May 3 '14 at 23:55

Hint: Replace $f_n$ by a seuqence converging to $f$, that is also dominated by $f$. For example take $g_n= \min (f_n,f)= \frac{f_n+f-|f_n-f|}{2}$.

Then use dominated convergence, and the fact that $f,f_n\geq 0$.

Your goal is to show that $\|f_n-f\|_1\to 0$, and then the claim will follow.

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Thanks, I see that you use the norm of the functions in this solution. I haven't learned about that type of norm yet. I will try to understand this solution when I learn about the L1(?) norm. – user119615 May 4 '14 at 0:11
$\|f_n-f\|= \int |f_n-f| d\mu$. So if this converges to $0$, then $$|\int_E f_nd\mu -\int_E fd\mu| \leq \int_E |f_n-f|d\mu \leq \int_X |f_n-f|d\mu \to 0$$ – Dimitrios Nt May 4 '14 at 0:54
I see. Do you know if you are able to solve it if you have only learned how to integrate non-negative functions, and you have not learned the dominated convergence theorem? – user119615 May 4 '14 at 15:39
I read about what you did. And I got from DCT: $lim_{n \rightarrow \infty}\int(f_n+f+|f_n-f|)/2d\mu=\int fd\mu$. If I just take the integral over every function alone on the LS. I get that $lim_{n \rightarrow \infty}\int|f_n-f|d\mu=0.$ But the problem is, we know that we if we have an integral over many functions we can take the integral over each one. But when it is combined with the limit, is it then always clear that we can just split it up? – user119615 May 4 '14 at 18:13
Yes you can split it because the integrals $\int f_n ,\int f$ exist, and also $\lim_{n\to\infty}\int f_n$ is finite. There is not the problem of getting $\infty- \infty$. Also if you don't want to use DCT, then using two times Fatou's lemma is actually the proof of DCT. – Dimitrios Nt May 4 '14 at 18:21

You use the part about infinity to say that $$\int_E fd\mu =\int f d\mu - \int_{E^C} f d\mu$$ (the same holds for the $f_n$ similarly, at least from a certain index) and this, repeating the same argument with fatou as you did in the question with $E^C$, gives you

$$\liminf_{n\to\infty} \int_{E^C} f_n d\mu \geq \int_{E^C} f d\mu$$

then you get

$$\liminf_{n\to\infty} \int_{E} f_n d\mu= \int f d\mu - \liminf_{n\to\infty} \int_{E^C} f_n d\mu \leq \int f d\mu - \int_{E^C} f d\mu = \int_{E} f d\mu.$$

I'm a bit rusty on the properties of liminf, does this convince you?

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Thank you very much! – user119615 May 4 '14 at 0:10
I just saw something that might be a problem in the first equality in the last line. $lim inf_{n \rightarrow \infty}(a_n+b_n) \ge lim inf_{n \rightarrow \infty} a_n+liminf_{n \rightarrow \infty} b_n$. Have you used equality there`? – user119615 May 4 '14 at 0:17
i have, and it's wrong in general. anyway it's not a problem, the inequality is in the right direction isn't it? – Franco May 4 '14 at 0:27
I see, this is what you did?: $\int f d\mu=lim \int f_n d\mu=\liminf \int f_n d\mu=\liminf(\int_Ef d\mu +\int_{E^C})f_n d\mu \ge\liminf \int_E fd\mu+\liminf \int_{E^C}fd \mu$. So this give that $\liminf \int _Ef d\mu \le \int f d\mu -\int_{E^C} d\mu$ ? And when the lim inf both is larger and smaller than a given number, the limit must exist and be that number, so hence you proved existence and what the limit is? – user119615 May 4 '14 at 0:36
exactly, combining my computation with the inequality you have in your attempt you get the existence and the value of the limit – Franco May 4 '14 at 0:49