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I'm trying to prove the Monotone Convergence Theorem for decreasing sequences, namely if

Let $(X,\mathcal{M},\mu)$ be a measure space and suppose $\{f_n\}$ are non-negative measurable functions decreasing pointwise to $f$. Suppose also that $f_1 \in \mathscr{L}(\mu)$. Then $$\int_X f~d\mu = \lim_{n\to\infty}\int_X f_n~d\mu.$$

Why does this statement not follow from LDCT with $f_n$ being dominated by $f_1$?

I'm also aware of the solutions with $g_n=f_1-f_n$, but the question asks to prove it using Fatou's lemma

Anthony Peter
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  • In Fatou's lemma, the inequality goes the other way. So the inequality $\lim\int f_n,d\mu\le\int f,d\mu$ doesn't follow from Fatou. – grand_chat Oct 09 '15 at 01:26
  • @grand_chat why does the statement not follow from LDCT with $g=f_1$? – Anthony Peter Oct 09 '15 at 01:59
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    I think it does follow from LDCT. –  Oct 09 '15 at 03:01
  • @JohnMa Then why does every solution use the $g_n=f_1-f_n$ trick and MCT? – Anthony Peter Oct 09 '15 at 03:03
  • I guess that's because LDCT is an overkill? (It seems that LDCT is the last theorem to prove, either Fatou or MCT comes first) –  Oct 09 '15 at 03:05
  • @JohnMa That's peculiar, since the problem is given at the end of ch 1 of big rudin, which proves LDCT before the problems are posed. – Anthony Peter Oct 09 '15 at 03:18
  • Certainly it follows from DCT. Why do you say every solution uses that $g_n=f_1-f_n$ thing instead? I mean have you seen every solution? It's not like there's an official solution set somewhere... – David C. Ullrich Oct 09 '15 at 13:30

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As you say in the comments, the problem is posed at the end of the first chapter of Rudin's Real and Complex Analysis, so it is not clear whether Rudin intends the student to use the Monotone Convergence Theorem, Fatou's Lemma or the Dominated Convergence Theorem.

You are right that the proof is a trivial application of DCT. However, it is certainly an instructive exercise to prove it using just MCT or Fatou's Lemma.