Let $f_n : [0,1]\to \mathbb R$ be a sequence of Riemann integrable functions. We call $f$ Riemann integrable if $U(f) = \inf_P U(f,P) = \sup_P L(f,P) = L(f)$ where $U(f,P)$ denotes the upper sum with respect to partition $P$ and $L(f,P)$ the lower sum.

If $f_n \to 0$ pointwise then it is possible to produce an example of $f_n$ such that $\int f_n$ is unbounded. It is even possible with $f_n$ continuous. But I want to show that it's not possible anymore if $f_n$ are uniformly bounded (I'm quite sure this is true). I tried to prove my conjecture but failed and I'm not sure what the problem is.

Let $f_n \to 0$ pointwise and let $|f_n|\le M$ for some $M > 0$. Let $\varepsilon > 0$. Then the goal is to show that for $n$ greater some $N$:

$$ \int_0^1 f_n \le \varepsilon$$

The definition of Riemann integral lets us use either upper or lower sum instead: If for all partitions $P$ we have $U(f,P) \le \varepsilon$ then $U(f) \le \varepsilon$. But this doesn't seem to be helpful here since we have no information about $f_n$.

How to prove that if $f_n \to 0$ pointwise and $|f_n|\le M$ then $\int f_n \to 0$?

  • 3
    $\begingroup$ See this for an elementary proof (i.e., no measure theory). $\endgroup$ – David Mitra Jul 8 '14 at 19:22
  • $\begingroup$ @DavidMitra Thank you. This fully answers my question. Can you post your comment as an answer? $\endgroup$ – blue Jul 9 '14 at 7:12

This is difficult to do without using measure theory! The result is known as Arzela's Bounded Convergence Theorem.

An elementary proof proof can be found here.

| cite | improve this answer | |
  • $\begingroup$ Here is an updated link. $\endgroup$ – David Mitra Oct 25 '19 at 14:24

It is straightforward from the Lebesgue Dominated Convergence Theorem and the fact that any Riemannian integrable function is a Lebesgue integrable function on $[0,1]$.

| cite | improve this answer | |

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.