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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.

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    $\begingroup$ @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 $[0,1]$. $\endgroup$ – Guangyao Dec 5 '20 at 22:25
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    $\begingroup$ @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. $\endgroup$ – Guangyao Dec 5 '20 at 22:30
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    $\begingroup$ 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 $\alpha\to 0$. $\endgroup$ – Stinking Bishop Dec 5 '20 at 22:33
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    $\begingroup$ @Guangyao: so as you want. $\endgroup$ – Paresseux Nguyen Dec 5 '20 at 22:52
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    $\begingroup$ @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. $\endgroup$ – RRL Dec 5 '20 at 23:36
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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 $\int_{0}^{1}g(x^{n})dx\rightarrow g(0)$.

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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 $\alpha\to 0$.

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  • $\begingroup$ Thanks! Would you be able to provide a full answer which includes convergence of the first integral, for reference and completeness? $\endgroup$ – Guangyao Dec 5 '20 at 22:59
  • $\begingroup$ @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: math.stackexchange.com/a/571854/700480 . $\endgroup$ – Stinking Bishop Dec 5 '20 at 23:05
  • $\begingroup$ @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. $\endgroup$ – Paresseux Nguyen Dec 5 '20 at 23:10
  • $\begingroup$ @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 $\endgroup$ – Guangyao Dec 5 '20 at 23:14
  • $\begingroup$ @ParesseuxNguyen Can you point where, in the accepted answer, they use continuity at any other point than $0$? $\endgroup$ – Stinking Bishop Dec 5 '20 at 23:19
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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.
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