TITLE: Bepo - levi implies monotonone convergence theorem??
QUESTION [5 upvotes]: For Lebesgue measurable functions, Is is true that Beppo-Levi theorem implies the monotone convergence theorem?
Beppo-Levi: Suppose $ \sum_{k=1}^{\infty} \int |f_k |\, dm $ is finite. Then the series $\sum f_k(x) $ converges for almost all $x$, its sum is integrable and 
$$ \int \sum f_k \,dm = \sum \int f_k\, dm $$
Monotone convergence theorem: if $\{ f_n \}$ is a sequence of non-negative measurable functions and $\{ f_n(x) : n \geq 1 \}$ increases monotonically to $f(x)$ for each $x$ $( f_n \to_{pointwise} f $), then $$ \lim \int\limits_E f_n\, dm = \int\limits_E f\, dm $$

REPLY [5 votes]: The hypothesis for Beppo-Levi can be changed as $f_k$ being non-negative. It is equivalent of Tonelli theorem. Now,
let $g_k(x) = f_{k+1}(x)- f_k(x)$ and $f_0(x) = 0 $. Then
$$f_n(x) = \sum_{k=0}^{k=n-1}g_k(x)$$ and each $g_k(x) \geq 0$ because of monotonicity of $f_k$. So applying Beppo Levi on $g_k$ we have that 
$$  \int \sum_{k=1}^\infty g_k dm = \sum_{k=1}^\infty \int g_k dm $$
$$\sum_{k=1}^\infty g_k = \lim_{n \to \infty}\sum_{k=1}^n g_k = \lim_{n \to \infty} f_{n+1} = \lim_{n \to \infty} f_{n} $$ and
$$\sum_{k=1}^\infty \int g_k dm = \lim_{n \to \infty}\sum_{k=1}^n \int g_k dm = \lim_{n \to \infty} \int f_{n+1} dm = \lim_{n \to \infty} \int f_{n} dm$$ and hence we have monotone convergence theorem.