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Important Notice: Observe that although this question is very similar to many other questions such as the one in proving $\lim\limits_{n\to\infty} \int_{0}^{1} f(x^n)dx = f(0)$ when f is continous on [0,1], this problem only assumes continuity at $0$, but not on $[0,1]$.

Here is the question: Assume $g$ is (Riemann) integrable on $[0, 1]$ and continuous at $0$. Show $\lim_{n \to \infty} \int_0^1 g(x^n) dx = g(0)$.

My attempt at this question: I split up the integral into two parts from $[0,1-\alpha]$ and $[1-\alpha,1]$. For $[0,1-\alpha]$, there is uniform continuity, so the limit $n \to \infty$ can be shifted inside the integral to get $g(0)$.

But how do I keep the integral over $[1-\alpha,1]$ to be small?

Note: This is Exercise $7.4.10$ in Abbott, Understanding Analysis, 2nd edition.

Guangyao
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  • If you did notice the uniform continuity, I don't think the rest would be a difficulty for you, as you see $g$ is bounded on $[0,1]$ so you can have any estimation you want. – Paresseux Nguyen Dec 05 '20 at 22:24
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    @ParesseuxNguyen It is given that $g$ is continuous only at $0$ and not on the whole interval $[0,1]$. From what I know, only if $g$ is continuous on the compact interval $[0,1]$, then we can claim that $g$ is uniformly continuous on $[0,1]$. – Guangyao Dec 05 '20 at 22:25
  • Ah okay, in this case, you can use the integrability – Paresseux Nguyen Dec 05 '20 at 22:26
  • I do admit that I did not carefully read your post, it's a little bit trickier than its related version when $g$ is continuous. I'll post an answer to your post. – Paresseux Nguyen Dec 05 '20 at 22:27
  • @ParesseuxNguyen appreciate your comment(s), but can you help to provide a complete answer/response? – Guangyao Dec 05 '20 at 22:28
  • Is $g$ assumed to be Riemann- or Lebesgue-integrable? –  Dec 05 '20 at 22:28
  • Doesn't "integrable" imply "bounded" in this context [proper riemann integration]? – Brian Moehring Dec 05 '20 at 22:28
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    @StinkingBishop In this chapter of Abbott, only Riemann integration is defined and Lebesgue integration was not introduced yet. Hence, only definitions/concepts related to Riemann integration can be applied. – Guangyao Dec 05 '20 at 22:30
  • @BrianMoehring Suppose $g$ is bounded, I don't see how that helps in getting the complete answer. Would definitely appreciate if you can post a complete answer. – Guangyao Dec 05 '20 at 22:31
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    Then $g$ is bounded and $|\int_{1-\alpha}^1 g(x^n)dx|\le\alpha\sup_{0\le x\le 1}|g(x)|\to 0$ as $\alpha\to 0$. –  Dec 05 '20 at 22:33
  • I would just be repeating Stinking Bishop's comment. In order, choose $\epsilon > 0$. Then choose $\alpha \in (0,\epsilon)$ so that the integral on $[1-\alpha, 1]$ is sufficiently small for every $n$ (justified by StinkingBishop's comment). Then choose $N$ so that $g(x^n)$ is sufficiently close to $g(0)$ for all $n > N$ and $0 \leq x \leq 1-\alpha$. – Brian Moehring Dec 05 '20 at 22:42
  • The claim that $g$ is R-integrable implies $g$ is bounded is not true. – Zhanxiong Dec 05 '20 at 22:46
  • @Zhanxiong You need the function to be bounded in order to be able to define Darboux sums in the first place. –  Dec 05 '20 at 22:47
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    @Guangyao: so as you want. – Paresseux Nguyen Dec 05 '20 at 22:52
  • @Zhanxiong https://math.stackexchange.com/questions/610054/if-a-function-fx-is-riemann-integrable-on-a-b-is-fx-bounded-on-a – Brian Moehring Dec 05 '20 at 22:58
  • I am agreeing with @BrianMoehring on boundedness because the definition of Riemann integrability in the reference text (where this question is posted) assumes boundedness as well. – Guangyao Dec 05 '20 at 23:24
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    @BrianMoehring: The duplicate answer uses only that $g$ is bounded and continuous at $x = 0$. Since $x^n \to 0$ uniformly on $[1,1-\epsilon]$ there is $N(\epsilon)$ such that $n > N(\epsilon)$ implies $\int_0^{1-\epsilon} |g(x^n)-g(0)| , dx < \epsilon(1-\epsilon)$. I'll also vote to reopen if there is really a case made where all of this adds something new in addition to probably a dozen other duplicates or near duplicates of this question. The boundedness follows presumably from the riemann-integration tag. It seems the posts here already provide the OP with the required answer. – RRL Dec 05 '20 at 23:36
  • @RRL I edited the question to explicitly state Riemann integrability. This is implicitly assumed in the reference textbook because of the previous sections in it. Also, I think you made a typo in your reply above, Do you mean $x^n \to 0$ uniformly on $[0,1-\epsilon]$? – Guangyao Dec 06 '20 at 00:09

3 Answers3

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The proof is straightforward if we assume Lebesgue integration theory: Since $g$ is Riemann integrable, it is bounded. Choose $M>0$ such that $|g(x)|\leq M$ for all $x\in[0,1]$. For each $x\in[0,1)$, $g(x^{n})\rightarrow g(0)$ as $n\rightarrow\infty$ because $g$ is continuous at $0$. Moreover, $|g(x^{n})|\leq M$. By Dominated Convergence Theorem, we have \begin{eqnarray*} & & \lim_{n\rightarrow \infty}\int_{0}^{1}g(x^{n})dx\\ & = & \int_{0}^{1}\lim_{n\rightarrow\infty}g(x^{n})dx\\ & = & \int_{0}^{1}g(0)dx\\ & = & g(0). \end{eqnarray*}

Alternative approach that does not invoke Lebesgue integration theory: Since $g$ is Riemann integrable, it is bounded. Choose $M>0$ such that $|g(x)|\leq M$ for all $x\in[0,1].$ Let $\varepsilon>0$ be given. Choose $\delta>0$ such that $|g(x)-g(0)|<\varepsilon$ whenever $x\in[0,\delta]$. Choose $a\in(0,1)$ such that $a>1-\varepsilon$ . Observe that $a^{n}\rightarrow0$, so there exists $N$ such that $a^{n}\in[0,\delta]$ whenever $n\geq N$. Note that for any $x\in[0,a]$, we have that $0\leq x^{n}\leq a^{n}$, so $x^{n}\in[0,\delta]$ whenever $n\geq N$ and $x\in[0,a]$. Consider \begin{eqnarray*} & & \left|\int_{0}^{1}g(x^{n})dx-g(0)\right|\\ & = & \left|\int_{0}^{1}g(x^{n})dx-\int_{0}^{1}g(0)dx\right|\\ & \leq & \int_{0}^{1}\left|g(x^{n})-g(0)\right|dx\\ & = & \int_{0}^{a}\left|g(x^{n})-g(0)\right|dx+\int_{a}^{1}\left|g(x^{n})-g(0)\right|dx. \end{eqnarray*} If $n\geq N$, and $x\in[0,a]$, we have $x^{n}\in[0,\delta]$, so $|g(x^{n})-g(0)|<\varepsilon.$ It follows that $\int_{0}^{a}\left|g(x^{n})-g(0)\right|dx\leq\int_{0}^{a}\varepsilon dx\leq\varepsilon$. On the other hand, \begin{eqnarray*} & & \int_{a}^{1}\left|g(x^{n})-g(0)\right|dx\\ & \leq & \int_{a}^{1}2Mdx\\ & = & 2M(1-a)\\ & \leq & 2\varepsilon M. \end{eqnarray*} That is, $\left|\int_{0}^{1}g(x^{n})dx-g(0)\right|\leq\varepsilon(2M+1)$ whenever $n\geq N$. This shows that $\int_{0}^{1}g(x^{n})dx\rightarrow g(0)$.

Danny Pak-Keung Chan
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1

If $g$ is assumed to be Riemann-integrable on $[0,1]$, then it is bounded on $[0,1]$, and we have:

$$\begin{array}{rcl}\left|\int_{1-\alpha}^1 g(x^n)dx\right|&\le&\int_{1-\alpha}^1|g(x^n)|dx\\&\le&\left(\sup_{0\le x\le 1}|g(x)|\right)\int_{1-\alpha}^1 dx\\&=&\alpha\sup_{0\le x\le 1}|g(x)|\\&\to&0\end{array}$$

as $\alpha\to 0$.

  • Thanks! Would you be able to provide a full answer which includes convergence of the first integral, for reference and completeness? – Guangyao Dec 05 '20 at 22:59
  • @Guangyao Having seen that this question has been marked as a duplicate, I don't believe I can add much to the accepted answer there: https://math.stackexchange.com/a/571854/700480 . –  Dec 05 '20 at 23:05
  • @StinkingBishop Hmm, I don't want to be involved but you know, in your link $f$ is assumed to be continuous on $[0,1]$ but in this case, it's only at $0$ the function $g$ is continuous. – Paresseux Nguyen Dec 05 '20 at 23:10
  • @StinkingBishop Yes, I would like to highlight that continuity is only at $0$, and not on $[0,1]$. Hence, it is not a duplicate. I added an additional note to explain this in the question. Hope you will be able to add more details to make the answer complete. Thanks – Guangyao Dec 05 '20 at 23:14
  • @ParesseuxNguyen Can you point where, in the accepted answer, they use continuity at any other point than $0$? –  Dec 05 '20 at 23:19
  • Well, if you see that as duplicate, so please go ahead. – Paresseux Nguyen Dec 05 '20 at 23:23
  • They use that $\sup_{x\in[0,1]} |g(x)| < +\infty$ which does not follow from integrability of $g$ on $[0,1]$. – mechanodroid Dec 05 '20 at 23:35
  • @mechanodroid We are apparently assuming Riemann-integrability here (see the discussion below this question). –  Dec 05 '20 at 23:37
  • Also note riemann-integration tag, but OP needs to be clear otherwise we can answer this with dominated convergence. – RRL Dec 05 '20 at 23:38
  • True, I'm just noting where was continuity of $g$ on $[0,1]$ used in the linked answer. – mechanodroid Dec 05 '20 at 23:39
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So here is a quick answer because I kinda wanna stay out of any possible discussion at the moment.
Answer
WLOG: $g(0)=0$
I'll start by giving some straight point: $$ \int_{0}^1 g(x^n)dx = \int_{0}^1 g(t) \frac{1}{n} t^{-1+1/n}dt$$ As $g$ is continuous at $0$, for any $\delta>0$, there is a $1>\epsilon>0$ such that: $$ |g(x)| \le \delta \quad \forall \quad |x| \le \epsilon $$ Thus $$ \left| \int_{0}^1 g(t) \frac{1}{n} t^{-1+1/n}dt \right| \le \underbrace{ \left( \int_{0}^{\epsilon} \delta\frac{1}{n} t^{-1+1/n}dt \right)}_{ \le \delta}+\underbrace{ \left( \int_{\epsilon}^1 |g(t)|\frac{1}{n} \epsilon^{-1+1/n}dt \right) }_{ \le \frac{1}{n}\epsilon^{-1} \int_{0}^1 |g(t)|dt}$$ Thus , $$ \limsup_n \left| \int_{0}^1 g(t) \frac{1}{n} t^{-1+1/n}dt \right| \le \delta$$ Hence the conclusion

Comment

  • This is okay whether the integrability is understoond in which sense, Riemann or Lesbeque.
  • The formula of changing variables in the very beginning is not really necessary. A similar argument can be constructed without it, but surely more details need to be handle. Fortunately, they are just technical. The idea stays the same.
Paresseux Nguyen
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