0
$\begingroup$

I'm learning about measure theory, specifically the Lebesgue integral of nonnegative functions, and need help with the following problem:

If $f, f_n: \mathbb{R} \to [0, +\infty)$ measurable, $f_n \to f$ pointwise on $\mathbb{R}$ and $f_n \leq f, \forall n \in \mathbb{N}$, show that $\int_{\mathbb{R}}f = \lim_{n \to \infty} \int_{\mathbb{R}}f_n$.

The assumptions are (I always rewrite the problem to see if I understand it correctly):

  1. $f$ and $f_n$ are nonnegative measurable functions.
  2. $\lim_{n \to \infty}f_n = f$ (pointwise limit).
  3. $f_n \leq f$ for all $n$.

If I could commute the limit and the integral then I would be done (but it is never that simple, unfortunately).


This problem looks like an application of Fatou's lemma, i.e.

If $f_n$ is a sequence of nonnegative functions, then $$\int \Big(\liminf_{n \to \infty} \ f_n \Big) \leq \liminf_{n \to \infty}\int f_n$$

but I'm having difficulities applying it to this specific problem.

$\endgroup$
  • 1
    $\begingroup$ I'd say that $g_n(x) = \sup_{m \le n} f_m(x)$ is ready for the monotone convergence theorem $\endgroup$ – reuns Apr 14 '16 at 12:11
  • $\begingroup$ @user1952009 since there exists a monotone subsequence that converges subsequence, monotone convergence theorem; then by the convergence of $f_n$, you can finish from there. $\endgroup$ – Andres Mejia Apr 14 '16 at 12:14
3
$\begingroup$

Your assumption about Fatou's lemma is correct. Note that, using your assumptions, the left hand sides equals $\int f$. Now, on the other hand, by monotonicity of the integral, $\int f_n\le \int f$ for every $n$. Can you complete the reasoning from here?

$\endgroup$
  • $\begingroup$ To ensure my reasoning is correct, when you say "the left hand side equals $\int f$" (in Fatou's lemma I presume) is this because $\lim_{n \to \infty}f_n = f \implies \liminf_{n \to \infty}f_n = f$? That is, the limit is "matching" with the $\liminf$, is that correct? I also understand that, because by assumption $f_n \leq f$ for all $n$, by monotonicity of the integral $\int f_n\le \int f$ for all $n$ but it is still unclear to me as to how to complete the reasoning from here. $\endgroup$ – Von Kar Apr 14 '16 at 12:30
  • $\begingroup$ @VonKar if the limit exists then the lim inf exists and equals the limit, yes (in the sense the convergence is meant, pointwise in this case). And I was referring to Fatous lemma, yes. Write down the definition of the limes inferior, then this should show you that the inequality $\int f_n \le \int f$ implies that the reverse inequality (compared to what Fatous lemma tells you) also has to hold (actually you get $\limsup \int f_n \le \int f$). $\endgroup$ – Thomas Apr 14 '16 at 12:38
  • $\begingroup$ The definition I am familiar for the limit inferior is $$\liminf_{n \to \infty} \ f_n = \lim_{n \to \infty} \big(\inf_{m \geq n}f_m\big)$$ but I don't see how this implies the reverse inequality $\limsup \int f_n \le \int f$. Given the above result, the way to conclude would be since $\liminf \int f_n \geq \int f$ and $\limsup \int f_n \le \int f$ then $\lim \int f_n = \int f$. Is that correct? The part that I need to understand is the reverse inequality. $\endgroup$ – Von Kar Apr 14 '16 at 14:02
  • 1
    $\begingroup$ @VonKar if $\int f_n \le \int f$ for every $n$ then $\limsup \int f_n \le \int f_n$, because of course $\sup_{m>n} \int f_m$ is then also bounded by $\int f$ from above. Of course you need to apply the definition of $\limsup$ for this. From Fatou you get $\int f \le \liminf f_n$, from the previous reasoning $\limsup \int f_n\le \int f$. Since the reverse inequality is always true you necessarily have equality. $\endgroup$ – Thomas Apr 14 '16 at 14:17
  • $\begingroup$ Thank you for your detailed and clear explanation (and your time). It was extremely helpful and understandable. $\endgroup$ – Von Kar Apr 14 '16 at 16:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.