# Given that $g$ is only continuous at $0$, not on $[0,1]$, show $\lim_{n \to \infty} \int_0^1 g(x^n) dx = g(0)$.

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.

• @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 $[0,1]$. Dec 5, 2020 at 22:25
• @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. Dec 5, 2020 at 22:30
• 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 5, 2020 at 22:33
• @Guangyao: so as you want. Dec 5, 2020 at 22:52
• @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 5, 2020 at 23:36

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 $$\int_{0}^{1}g(x^{n})dx\rightarrow g(0)$$.

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? Dec 5, 2020 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: math.stackexchange.com/a/571854/700480 . Dec 5, 2020 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. Dec 5, 2020 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 Dec 5, 2020 at 23:14
• @ParesseuxNguyen Can you point where, in the accepted answer, they use continuity at any other point than $0$? Dec 5, 2020 at 23:19

So here is a quick answer because I kinda wanna stay out of any possible discussion at the moment.
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