Bepo - levi implies monotonone convergence theorem??

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 $$\lim \int\limits_E f_n\, dm = \int\limits_E f\, dm$$

• Levi's theorem is usually used to prove Lebesgue's theorem, along with Fatou's lemma. – Pedro Tamaroff May 4 '14 at 22:25

The monotone convergence theorem handles infinities gracefully, which can only be done for functions that are positive (or otherwise reasonably controlled from below). In particular, nowhere does it assume that $f_n$, $f$, or the integrals, are finite. This seems to be beyond the scope of Beppo Levi, so I'm not sure that fixing this issue alone is considerably easier than proving everything from scratch. But let me try.

Depending on your version of definitions, it may or may not be trivial that for a positive function $f$ the following special case of monotone convergence holds:

$$\intop_E f dm = \lim_{C \to +\infty, E_n \uparrow E} \intop_{E_n} \min(f, C) dm$$

where $E_n$ are sets of finite measure that approximate $E$ (I assume $\sigma$-finiteness; if it fails then we should restrict to $\{f > 0\}$; if it fails even there then monotone convergence holds almost trivially with both sides infinite).

Now in order to make use of Beppo Levi we should make the limit finite. I would do that by replacing $f_n$ by $f_{n,C,k} := (f_n \wedge C) \mathsf{1}[E_k]$ and $f$ by $f_{C,k} := (f \wedge C) \mathsf{1}[E_k]$ for some fixed $k$. Now we can safely apply Beppo Levi to the successive differences $f_{n+1,C,k} - f_{n,C,k}$ to obtain

$$\intop f_{C,k} dm = \lim_{n \to \infty} \intop f_{n,C,k} dm$$

Now take $C$ and $k$ to $\infty$ and interchange the limits. You can always do this with monotonely increasing limits (this is equivalent to rearrangement of terms in positive series, or Fubini on $\mathbb{N} \times \mathbb{N}$, or whatever you prefer). On the other hand, monotone convergence itself is about rearrangement or Fubini on $E \times \mathbb{N}$, so I'm not even sure you would view the things that I rely on as more basic than those that you prove...

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.