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My notes report the following assertion for the theorem:

Beppo Levi's Theorem: Let $E$ be a measurable set and $\{ f_n(x)\}$ a sequence of integrable functions on E, such that $\lim\limits_{n\to\infty} f_n(x) = f(x)$ (pointwise convergence) almost everywhere on E, and $f_n(x)\leq f(x)$. Then $f(x)$ is integrable on E and $\lim\limits_{n\to\infty} \int\limits_E f_n(x) = \int\limits_E f(x)$

Is this correct? Cause my book reports multiple versions of the theorem, but not this one.

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  • $\begingroup$ What does punctual convergence mean? $\endgroup$
    – BigbearZzz
    Jan 16, 2019 at 15:17
  • $\begingroup$ Pointwise, my bad, translation error. Let me correct it $\endgroup$ Jan 16, 2019 at 15:18
  • $\begingroup$ It should be only $\lim\limits_{n\to\infty} f_n(x) = f(x)$ without "for every $n$" because $n$ is already in the limit sign. $\endgroup$
    – BigbearZzz
    Jan 16, 2019 at 15:20
  • $\begingroup$ I don't know why, but my notes report both. I'll take your tip anyway! $\endgroup$ Jan 16, 2019 at 15:21
  • $\begingroup$ As is, the statement is wrong, because $E=\mathbb{R}^+$, $f_n=1_{[-n,n]}$ seems to give your theorem a problem. $\endgroup$
    – Aphelli
    Jan 16, 2019 at 15:30

2 Answers 2

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First this theorem is wrong

Consider $E = \mathbb R$, $f=0$ and $f_n = -\chi_{[n,n+1]}$. Where $\chi_{[n,n+1]}$ is the indicator function of the interval $[n,n+1]$.

They satisfy the hypothesis, but the conclusion doesn't hold as

$$ -1 = \int_{\mathbb R} f_n \neq \int_{\mathbb R} f = 0$$ while $(f_n)$ converges pointwise to the always vanishing function $f$.

Second this would more related to dominated convergence theorem

See Dominated convergence theorem if the hypothesis would be correctly set.

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The theorem is usually known as the monotone convergence theorem and comes with an extra requirement that $$ f_1(x)\le f_2(x)\le\dots \le f(x) $$ for almost every $x$. Other than this the other parts of your statement is correct.

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  • $\begingroup$ My notes report that the classical statement requires that $f_1(x) < f_2(x) < \dots < f_n(x) < \dots$ but it can be proved not to be necessary... Isn't that right? $\endgroup$ Jan 16, 2019 at 15:26
  • $\begingroup$ That’s not the monotone convergence theorem. There is no assumption on the sequence $f_n$. $\endgroup$
    – Aphelli
    Jan 16, 2019 at 15:27
  • $\begingroup$ @Bafforasta Then you'd need extra assumptions. As it is now the statement is not true. $\endgroup$
    – BigbearZzz
    Jan 16, 2019 at 15:30
  • $\begingroup$ @Mindlack I know the statement in the OP is not the MCT that's why I said that it missed one assumption, i.e. monotonicity. The reason being the name Beppo Levi is associated with the MCT so I made the link. $\endgroup$
    – BigbearZzz
    Jan 16, 2019 at 15:35
  • $\begingroup$ @BigbearZzz I just found this post: math.stackexchange.com/questions/1177788/beppo-levis-theorem this looks pretty similar $\endgroup$ Jan 16, 2019 at 15:46

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