4
$\begingroup$

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!

$\endgroup$
3
  • $\begingroup$ $M$ is a measurable set? Is the sequence really non-decreasing? One can also proof the theorem for $f_1\geq f_2 \geq ... \geq 0$ and $\int f_1<\infty$. The monotonte function theorem says someting about the integrability of the limit functions. It is integrable if and only if $\int f_n \leq C$. It's always measurable since it's the pointwise (a.e) limit of a sequence of measurable functions $\endgroup$
    – Quickbeam2k1
    Apr 7, 2013 at 19:04
  • 1
    $\begingroup$ But I think your doubts are right. This sequence does not need to be a bounded real sequence. However the funny thing is, that the result is trivially true for $\int f_k \to \infty$. Hence she may assume that the sequence is bounded. But in my eyes she should mention that. $\endgroup$
    – Quickbeam2k1
    Apr 7, 2013 at 19:10
  • $\begingroup$ You already know that you're correct, but your last paragraph is particularly insightful too. The Monotone Convergence Theorem is fundamentally about nonnegative measurable functions (whose integrals take values in $[0,\infty]$), while the Dominated Convergence Theorem is about integrable functions (whose integrals take values in $(-\infty,\infty)$ or some other Banach space). Another pair of theorems like this is Tonelli's Theorem (about nonnegative measurable functions) and Fubini's Theorem (about integrable functions). $\endgroup$
    – Toby Bartels
    Feb 28, 2018 at 12:35

1 Answer 1

Reset to default
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ I agree, that the equality of the two integrals is correct even in this case, i.e if the integral of $f$ is infinite. BUT the statement "then $f$ is integrable" in the theorem is incorrect. Right? $\endgroup$
    – harajm
    Apr 7, 2013 at 19:54
  • 1
    $\begingroup$ Sorry I didn't notice that. Yes you are right. The statement "f is integrable' in the conclusion is false. It cannot be concluded without further hypothesis on the sequence. $\endgroup$
    – user71352
    Apr 7, 2013 at 21:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .