Show that: $\lim \limits_{n \rightarrow+\infty} \int_{0}^{1}{f(x^n)dx}=f(0)$ Supose that $f\colon [0,1] \rightarrow \mathbb R$ is a continuous function. Show that: $$\lim \limits_{n \rightarrow+\infty} \int_{0}^{1}{f(x^n)dx}=f(0)$$
I'm studying the Riemann Integral theory and I have not idea to how do this. I need some hint to start because I do not see how to apply the Riemann sum or other theory concept to be able to help me.
 A: Since $g(x)=f(x)-f(0)$ is continuous iff $f(x)$ is, and $g(0)=0$, we may assume that $f(0)=0$ in the first place. 
Hint: The idea is that if $f$ is continuous at $x=0$ and $f(0)=0$; then $f_n(x)=f(x^n)$ goes to $0$ uniformly over $[0,\eta]$ for any $1>\eta>0$. Thus, given any $\varepsilon>0$; it suffices to take $\eta$ sufficiently close to $1$ to make the integral over $[\eta,1]$ tiny, and use uniform convergence to kill of the integral over $[0,\eta]$.
SPOILERS
Given $\varepsilon>0$, there exists $\delta>0$ such that $0<t<\delta\implies |f(t)|<\varepsilon$. 
Let $M>0$ be such that $|f(x)|<M$ over $[0,1]$. Given this $\delta>0$, choose $1>\eta>0$ such that $M(1-\eta)<\varepsilon$, and choose $N$ such that  $0\leq x\leq \eta$ implies $0\leq x^N \leq\delta$ (the same will hold true for $n>N$). Then for $n>N$ we will have 
$$\left|\int_0^1f(x^n)dx\right|\leqslant \int_0^\eta |f(x^n)|dx+\int_\eta^1 |f(x^n)|dx\leqslant \eta \varepsilon+M(1-\eta)<(1+\eta)\varepsilon<2\varepsilon$$
ADD Note that by our set up, if $x\in[0,\eta]$, then $0\leq x^n\leq \delta$. This means that $|f(x^n)|<\varepsilon$ over $[0,\eta]$, so $$\int_0^\eta |f(x^n)|dx\leqslant \int_0^\eta \varepsilon dx=\eta\varepsilon$$
A: Here's a slightly alternative way. As already mentioned, if the fact is true for $f(0)=0$ it is true for any $f(0)\neq 0$, so assume $f(0)=0$. Do a change of variables $u=x^n$. Then $du = n x^{n-1} dx = n u^{1-1/n} dx$ and so
$$I_n = \int_0^1 \frac{1}{n} u^{1/n-1} f(u) du = \int_0^\epsilon \frac{1}{n} u^{1/n-1} f(u) du + \int_\epsilon^1 \frac{1}{n} u^{1/n-1} f(u) du.$$
The first term is bounded in absolute value by
$$\sup_{u \in [0, \epsilon]} |f(u)| \cdot \int_0^\epsilon \frac{1}{n} u^{1/n-1} du=\sup_{u \in [0, \epsilon]} |f(u)| \;\epsilon^{1/n}.
$$
For the second integral, $u^{1/n-1}\leq \epsilon^{1/n-1}$ on our region of integration, so the second integral is bounded in absolute value by 
$$\frac{\epsilon^{1/n-1}}{n} \int_\epsilon^1 |f(u)| du.$$
Using these bounds,
$$\limsup_{n \to \infty} |I_n| \leq \limsup_{n \to \infty} \left[ \sup_{u \in [0, \epsilon]} |f(u)| \;\epsilon^{1/n} +  \frac{\epsilon^{1/n-1}}{n} \int_\epsilon^1 |f(u)| du. \right] =\sup_{u \in [0, \epsilon]} |f(u)|+0.$$
By the continuity of $f$, we have $\sup_{u \in [0, \epsilon]} \to 0$ as $\epsilon \to 0$, so $\limsup_{n \to 0} |I_n| = 0$ and hence $\lim_{n \to 0} I_n = 0$.
I'm pretty sure this way you could beat this problem into the form of a mollifier, that is an approximation to identity, if you wanted. 
