# Uniformly bounded sequence of Riemann integrable functions

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?

• 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. Dec 23, 2019 at 5:47
• Then,I guess this result holds since R.I functions on a closed bounded interval is Lebesgue m'ble. Dec 23, 2019 at 5:55
• Yes, RI functions are also Lebesgue integrable and DCT can be applied. Dec 23, 2019 at 5:56
• Too bad I wasn't thinking about that. Thanks anyways. Dec 23, 2019 at 5:58
• 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 Jan 4, 2020 at 6:04

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 .

• +1 Arzela theorem can also be proved as discussed in this thread. Jan 4, 2020 at 5:51
• @ParamanandSingh: Thanks for adding that.
– RRL
Jan 5, 2020 at 5:20

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 $$\int_{[a,b]} f_n(x) dx \to 0$$.

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.

• "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)? Dec 23, 2019 at 6:13
• We learn Egorov way earlier than DCT, why not this has some virtue? Plus, the domain is finite, and even suites better for Egorov. Dec 23, 2019 at 6:14
• @SL_MathGuy, yes, that is the same saying. Dec 23, 2019 at 6:16
• 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. Dec 23, 2019 at 6:18
• @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. Dec 23, 2019 at 6:21