I need some clarifications about the Lebesgue Convergence theorems, I am using the book Beginning Functional Analysis by Karen Saxe and I believe that she has stated the monotone convergence theorem wrong and also made a mistake in the proof. Let $M$ be the set of Lebesgue measurable sets. She writes:
Theorem 3.13 Suppose that $A\in M$ and that $\{f_k\}_{k=1}^{\infty }$ is a sequence of measurable functions such that $0\leq f_1(x) \leq f_2(x)\leq \dots$ for almost all $x\in A$. Let $f$ be defined to be the pointwise limit, $f(x)=\lim_{k \to \infty} f_k(x)$, of this sequence. Then $f$ is integrable and $\lim_{k \to \infty} \int_{A}f_k dm=\int_{A}f dm$.
In the proof she uses the monotonicity of the Lebesgue integral to conclude that $\int_{A}f_1dm \leq \int_{A}f_2 dm\leq \dots \leq \int_{A}f dm$ and she states that $\{\int_{A}f_kdm\}_{k=1}^{\infty}$ is a bounded and non decreasing sequence of real numbers.
My objections are that this sequence may NOT be assumed to be bounded since $\int_{A}f dm$ may be infinite, in fact she has stated in the theorem that she will prove it to be finite. As I would guess, please correct me and help me to clarify this, the word integrable should be changed to measurable (so that the integral is defined, this is really a separate statement, i.e. limit of measurable functions are measurable) and she should remove the word bounded from the proof and instead let the limit of this sequence be finite or infinite, the rest of the proof would then stay valid.
Am I correct if I understand it like this: The monotone convergence theorem and Fatou's Lemma doesn't say anything about the integrability of the limit function only of the value of the integrals (I.e this value may be infinite). But the Lebesgue Dominated convergence theorem states sufficient conditions for the limit function to actually be integrable? All comments are helpful!
Best Answer
I think she has assumed that the monotone sequence of integrals is bounded since if it were not then by Fatou's Lemma the liminf of the monotone sequence of integals will have to be infinite. That is
$\infty=\int_{A}fdm\le\liminf_{n\rightarrow\infty}\int_{A}f_ndm\le\limsup_{n\rightarrow\infty}\int_{A}f_ndm$
So $\lim_{n\rightarrow\infty}\int_{A}f_ndm=\int_{A}fdm$.
Even though this simplification of the proof is allowed it is worth justifying. Your concern is of course correct.