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Below is the common statement of the Monotone Convergence Theorem.

Suppose $\{f_n\}$ is a sequence of non-negative measurable functions with $f_n(x) \leq f_{n+1}(x)$ a.e. and $\lim \limits_{n\rightarrow\infty} f_n(x) = f(x)$. Then $$\lim \limits_{n\rightarrow\infty} \int f_n = \int f$$

But proof requires only $f_n(x) \leq f(x)$ since it is sufficient to conclude that $\int f_n \leq \int f$ for all $n$ by monotonicity. So what is the value of the more restrictive statement than it could be (except that it leads to a nice name for the theorem).

Edit: Proof.

By Fatou's lemma $\int f \leq \lim \limits_{n\rightarrow\infty} \inf \int f_n$. Now, given $f_n(x) \leq f(x)$ it follows that $\int f_n \leq \int f$ for all $n$ by monotonicity. Hence $\lim \limits_{n\rightarrow\infty} \sup \int f_n \leq \int f$. Which combined with Fatou's lemma gives the desired result.

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    $\begingroup$ Which proof are you following? $\endgroup$ – user99914 Oct 11 '17 at 23:40
  • $\begingroup$ Huh? No correct proof (including the fact that the limit is specifically the integral of $f$) can use just the inequality you showed. $\endgroup$ – mathguy Oct 11 '17 at 23:44
  • $\begingroup$ @JohnMa Added proof to the question. $\endgroup$ – user2734153 Oct 12 '17 at 0:05
  • $\begingroup$ @user2734153 Please check out the answer I provided. I think you will find it interesting. $\endgroup$ – LucasSilva Feb 18 '18 at 23:15
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The monotonicity condition is indeed redundant. I have two things I want to say about that.

  1. You gave a proof based on Fatou's lemma. So how do you prove Fatou's lemma? The standard way to prove Fatou's lemma is to use the monotone convergence theorem. And the monotone convergence theorem is proved using the definition of the integral as the sup of integrals of simple functions, the countable additivity of measures, and clever choice of sets (see Folland's textbook or either of Rudin's textbooks for details). It's possible to give a "direct" proof of Fatou's lemma by writing down the proof of the monotone convergence theorem I just described with $f_n$ replaced by $g_n = \sup_{n \geq k} f_k$ (which is monotone increasing even when $f_n$ is not) and using that $\liminf f_n = \lim_n g_n$, but that's really just the monotone convergence theorem in disguise.

  2. Your convergence theorem where the functions converge from below but not necessarily monotonically has been called the Convergence from Below Theorem. It is possible to give a direct proof of it. Basically, you just rework the direct proof of Fatou's lemma I described above (which was itself a reworking of the usual proof of the monotone convergence theorem) while being carefully not to write down $\liminf$. See the article "Convergence from Below Suffices" by JOEL F. FEINSTEIN https://www.maths.tcd.ie/pub/ims/bull59/M5902.pdf

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I guess that relaxing the hipotesis about monotonicity you get a weakening of the dominated convergence theorem (for nonnegative functions dominated by your $f(x)$). Here you construct the limit (by monotonicity) and you prove it's a measurable function. Hope this was useful.

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