Reading through Rudin's Real and Complex Analysis, I came across the following exercise:

Suppose $(f_n: X \to [0,\infty])$ is a monotone decreasing sequence of measurable functions such that $\lim\limits_{n \to \infty} f_n(x) = f(x)$ for all $x \in X$. Prove that if $f_1 \in L^1(\mu)$, then

$$\lim\limits_{n \to \infty}\int\limits_{X} f_n \, \mathrm{d}\mu = \int\limits_X f \, \mathrm{d}\mu.$$

It seems like this should be a trivial application of the Dominated Convergence Theorem, taking $f_1$ to be the dominating function. But it seems like an exercise would not be so trivial as to merit basically a one line proof. Is there a reason that DCT fails to be applicable here?

  • $\begingroup$ I presume the exercise was meant to apply the monotone convergence theorem to $f_1-f_n$? Integrability just ensures that you can conclude $\int f_n \to \int f$ from $\int (f_1-f_n) \to \int (f_1-f)$. $\endgroup$
    – copper.hat
    Jun 22 '13 at 17:57
  • $\begingroup$ The exercise is listed after the DCT has been used somewhat at length, or I would think that the purpose was to motivate the theorem. Also taking the sequence $f_1-f_n$ is essentially replicating the general proof of the DCT. $\endgroup$
    – Cameron
    Jun 22 '13 at 17:59
  • $\begingroup$ DCT is certainly applicable. After all, $0 \leq f_n \leq f_1$ for all $n$. You should try to find a few examples showing that the result can fail if $f_1$ is not assumed to be integrable. $\endgroup$
    – Martin
    Jun 22 '13 at 21:07

You are right, it is a trivial application of the Dominated Convergence Theorem. Since the problem is listed at the end of the first chapter, it is not clear whether Rudin intends the student to use the Dominated Convergence Theorem or the Monotone Convergence Theorem. It is instructive to complete the exercise by applying each of the theorems.

You can check the following question for a proof that uses the Monotone Convergence Theorem: Monotone Convergence Theorem for non-negative decreasing sequence of measurable functions


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.