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I'm studying this proof:

https://www.math.utah.edu/~savin/L2_5210.pdf

but I can't understand the step when he says: Hence, by the Monotone Convergence Theorem, there exists an integrable function $\phi$, such that $\lim_{n\to \infty}\phi_n(x) =\phi(x)$ for almost all $x$.

The monotone convergence theorem that I have studied does not have the existence of the limit function $\phi$ in the thesis, but in the hypothesis. In fact, why the limit $\lim_{n\to \infty}\phi_n(x)$ could not be infinite on a non-zero measure set?

Nameless
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  • I believe this is covered since $\sum_{i}f_i$ is assumed to be absolutely convergent? – K.defaoite Dec 28 '21 at 17:34
  • I think the existence part follows directly (without the MCT), but proving that this limit is also integrable requires the MCT. – Zuy Dec 28 '21 at 18:07

1 Answers1

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Here is a way you can understand this step :

  1. $\varphi := \lim_{n \rightarrow +\infty} \varphi_n$ exists since $(\varphi_n)$ is an increasing sequence of function. At this step, we still don't know that $\varphi$ is integrable.

  2. By Monotone Convergence Theorem, $$\lim_{n \rightarrow +\infty} \int \varphi_n = \int \varphi$$

  3. Because you have $\displaystyle{\int \varphi_n \leq \sum ||f_i||}$ for every $n$, then you deduce from 2. that $$\int \varphi \leq \sum ||f_i|| < \infty$$

so $\varphi$ is integrable and you are done.

TheSilverDoe
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  • This method is clear if the monotone convergence theorem still holds for “extended” functions. That is, functions that could assume $+\infty$ as a value. Is it really this the case? Can one recycle the proof of the MCT also for this type of functions, without variations? – Nameless Dec 29 '21 at 09:12
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    @Nameless Yes. In Lebesgue theory, functions are generally allowed to take $\infty$ as a value, and all the main theorems remain true. You can check the proof of MCT to see that it has no influence. – TheSilverDoe Dec 29 '21 at 11:18