In the Dominated Convergence Theorem, we usually assume that $|f_n| \le g$ for some integrable function $g$. However, what is a counter-example where $f_n$ are not dominated by an integrable function but only by their pointwise limit itself, $f$?

That is, $|f_n| \le f$ for all $n$, and $f$ is merely measurable, and the limit of the sequence $(f_n)$ of measurable functions?

EDIT: Or does one not exist?


Take $g_n = |g_n| = n\Bbb 1_{\left(\frac{1}{n+1},\frac{1}{n}\right)}$. Since the supports of these functions intersect no where, with $f_n = |f_n|= \sum_1^k g_k$, the best dominating function is the point wise limit

$$f = \sum_{n=1}^∞ n \Bbb 1_{\left(\frac{1}{n+1},\frac{1}{n}\right)} $$ which is not (lebesgue) integrable, $$∫_ℝ f = ∫_0^1 f = \sum_{n=1}^{∞}\int_{\frac{1}{n+1}}^\frac{1}{n}n = ∞$$

A more easily defined family that has the same behavior is $f_n = \frac{\Bbb 1_{|x|>1/n}}{|x|}$.

  • $\begingroup$ $f$ is not the pointwise limit of $f_n$? $\endgroup$ – Roki Bak Sep 27 '15 at 21:08
  • $\begingroup$ Oh. Well one can easily take the partial sums instead to be $f_n$s. @RokiBak $\endgroup$ – Calvin Khor Sep 27 '15 at 21:17
  • $\begingroup$ But the integrals coincide? The limit of the integral of $f_n$ is $\infty$ too, so the equality in Dominated Convergence Theorem actually holds, since it just says $\infty = \infty$? $\endgroup$ – Roki Bak Sep 27 '15 at 21:31
  • $\begingroup$ Oh. It wasn't clear to me that you did wanted to violate the conclusion of DCT. If you are asking for the dominating function to be the limit as well, then the limit is non-negative, and we may without loss assume that each $f_n$ is non-negative, because we need $f_n(x)\uparrow f(x)\ge 0$. In such a situation the only divergence possible is to infinity. $\endgroup$ – Calvin Khor Sep 27 '15 at 21:38
  • $\begingroup$ @RokiBak sorry, forgot to tag you. (In addition, part of the conclusion of DCT is that the limit is an $L^1$ function.) $\endgroup$ – Calvin Khor Sep 27 '15 at 23:11

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.