4
$\begingroup$

Let {$f_n$} be a uniformly bounded sequence of Riemann int'ble functions on $[a,b]$.If $f_n\rightarrow 0$ pointwise then does it follow that $\int _{[a,b]}f_n\rightarrow0$?

My thoughts: The result doesn't follow from the given assumptions. To justify my claim, I choose $f_n(x)=\frac{x^2}{x^2+(1-nx)^2}$ on $[0,1]$ which satisfies all the criteria. Clearly, $f_n\rightarrow 0$ pointwise but I haven't been able to show that $\int _{[a,b]}f_n$ doesn't converge to $0$ although it's clear that it doesn't.

Are there any other counter-examples to justify this result?. I came up with $f_n(x)=nx(1-x^2)^n$ on $[0,1]$ but this choice of function doesn't have the uniform boundedness. Can anybody provide me with a relatively easy example to go with?

$\endgroup$
  • 3
    $\begingroup$ The intergal does tend to $0$. This is an easy consequence of DCT (Dominated Convergence Theorem) but I don't have a proof without measure theory. $\endgroup$ – Kavi Rama Murthy Dec 23 '19 at 5:47
  • $\begingroup$ Then,I guess this result holds since R.I functions on a closed bounded interval is Lebesgue m'ble. $\endgroup$ – SL_MathGuy Dec 23 '19 at 5:55
  • $\begingroup$ Yes, RI functions are also Lebesgue integrable and DCT can be applied. $\endgroup$ – Kavi Rama Murthy Dec 23 '19 at 5:56
  • $\begingroup$ Too bad I wasn't thinking about that. Thanks anyways. $\endgroup$ – SL_MathGuy Dec 23 '19 at 5:58
  • 1
    $\begingroup$ Your theorem is equivalent to the theorem of Arzela, see math.stackexchange.com/q/3039030/72031. There is a proof with minimal use of measure theory if the sequence $f_n$ is decreasing. See math.stackexchange.com/a/3038925/72031 $\endgroup$ – Paramanand Singh Jan 4 at 6:04
4
$\begingroup$

Arzela's bounded convergence theorem (1885) states that if $(f_n)$ is a uniformly bounded sequence of Riemann integrable functions on $[a,b]$ that converges pointwise to a Riemann integrable function $f$, then $\int_a^b f_n(x) \, dx \to \int_a^b f(x) \, dx$.

In this case $f = 0$ is Riemann integrable and $\int_a^b f_n(x) \, dx \to 0$ follows.

The assumptions are stronger than in convergence theorems for Lebesgue integrals since the Riemann integrability of the limit function is imposed. This can be proved without measure theory using elementary techniques, for example here .

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ +1 Arzela theorem can also be proved as discussed in this thread. $\endgroup$ – Paramanand Singh Jan 4 at 5:51
  • $\begingroup$ @ParamanandSingh: Thanks for adding that. $\endgroup$ – RRL Jan 5 at 5:20
3
$\begingroup$

As an immediate consequence of DCT and the fact that RI functions are also Lebesegue integrable (with the same value for the integral) we have $\int_{[a,b]} f_n(x) dx \to 0$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Perhaps you can do it in this way, still measure-theoretic: By Egorov, given $\epsilon>0$, some measurable set $S\subseteq [0,1]$ is such that $f_{n}\rightarrow 0$ uniformly on $[0,1]-S$ and $|S|<\epsilon$, then $\left|\displaystyle\int f_{n}\right|=\left|\displaystyle\int_{[0,1]-S}f_{n}+\int_{S}f_{n}\right|\leq\left|\displaystyle\int_{[0,1]-S}f_{n}\right|+\sup_{n}|f_{n}(x)||S|\leq\left|\displaystyle\int_{[0,1]-S}f_{n}\right|+\left(\sup_{n}|f_{n}(x)|\right)\cdot\epsilon$. We know that $\displaystyle\int_{[0,1]-S}f_{n}\rightarrow 0$ by the uniform convergence of $f_{n}\rightarrow 0$ on $[0,1]-S$, so $\left|\displaystyle\int f_{n}\right|$ is arbitrarily small.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ "given $ϵ>0$, some measurable set $S⊆[0,1]$ is such that $f_n→0$ uniformly on $[0,1]−S$ and $|S|<ϵ$" Is this equivalent to saying that $f_n \rightarrow0$ almost uniformly (which is the end result in Egorov's theorem)? $\endgroup$ – SL_MathGuy Dec 23 '19 at 6:13
  • $\begingroup$ We learn Egorov way earlier than DCT, why not this has some virtue? Plus, the domain is finite, and even suites better for Egorov. $\endgroup$ – user284331 Dec 23 '19 at 6:14
  • $\begingroup$ @SL_MathGuy, yes, that is the same saying. $\endgroup$ – user284331 Dec 23 '19 at 6:16
  • $\begingroup$ I learnt Egorovs after DCT. As you mentioned, the theorem can be applied since the measure space is finite and more importantly, I missed the observation that {$f_n$} is a sequence of measurable functions, which really is the key point here. $\endgroup$ – SL_MathGuy Dec 23 '19 at 6:18
  • $\begingroup$ @KaviRamaMurthy, you said that DCT is better known than Egorov, that's just a matter of your taste. Plus, I was using the textbook by J Yeh, and I learned Egorov way earlier than DCT. $\endgroup$ – user284331 Dec 23 '19 at 6:21

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.