Beppo Levi's theorem, is this assertion correct?

Question

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.

Answer 1

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.

Answer 2

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.