How to compute the limit of this integral? Let $f:[0,1]\rightarrow\mathbb{R}$ be a continuous function such that $f(0)\neq f(1)$. Define $f_n(x)=f(x^n)$ then I want to prove that $\lim_{n\rightarrow\infty}\int^1_0 f_n(x)dx=f(0)$.
Now, you can easily prove that the convergence is not uniform, so we can't switch the sign of limit and the sign of the integral. I tried to do a bunch of things to solve this problem, for example if $S(f_n,\sigma_N)$ is the upper sum of $f_n$ with respect to the equispaced partition $\sigma_N$ then i tried to prove that $\lim_{n\rightarrow\infty}\lim_{N\rightarrow\infty}S(f_n,\sigma_n)=f(0)$, this is true if we can switch the limits, but I don't know why we can. Could you help me please?
 A: Divide the interval in two parts:
A small interval near 1 where the integral is small because it is the length of the intervall times a bound of the function.
A large interval where you can bound the value of the function to be near $f(0)$ for large $n$.
A: It depends on what you can assume.
If you know Lebesgue integration then it is easy consequence of Dominated Convergence Theorem.
If you do not know Lebesgue Theory then you can prove DCT directly without "measure things". Of course it is much more than you ask but also much more general and with quite simple proof.

Theorem. Let $(f_n)$ be a sequence of Riemann integrable functions on $[0,1]$ such that $|f_n(x)| < M$ for $x \in [0,1]$ and $f_n(x) \to f(x)$ on $[0,1]$. Then
  $$
\lim_{n \to \infty} \int_0^1 f_n(x) dx = \int_0^1 \lim_{n \to \infty} f_n(x) dx.
$$

Proof. By linearity of integral we can assume that $f \equiv 0$ and $f_n(x) > 0$. Then we need to show that $\int_0^1 f_n(x) dx \to 0$ as $n \to \infty$. Assume that it is not true. 
Let $F_n := f_n/M$. Then $|F_n(x)| < 1$, $F_n(x) \to 0$ for $n \to \infty$ and by the assumption $I := \inf_n \int_0^1 F_n(x) dx > 0$ (here we choose suitable subsequence of $(F_n)$).
Let $s_n$ be a lower Riemann sum of $F_n$ such that $s_n \geq \frac{3}{4}I$. Than if $u_n$ is the sum of lengths of intervals of the partition of $[0,1]$ such that $F_n(x) > \frac{1}{2} I$ (on that intervals) then by $F_n(x) < 1$ and the definition of a lower sum we obtain $$\frac{3}{4} I \leq s_n \leq u_n + \frac{1}{2} I (1 - u_n) = \frac{1}{2} I + u_n (1-\frac{1}{2} I). $$
Hence $u_n \geq \frac{1}{4 I (1-I/2)} > 0$.
Let $U_n$ be the set of interiors of intervals defined by $u_n$. Define $V_n = \bigcup_{k=n}^\infty U_n$. Then each $V_n$ is open and the sequence $(V_n)$ is decreasing with $\inf_n |V_n| > 0$. Hence $\bigcap_n V_n \neq \emptyset$ and there exists $x_0$ such that $x_0 \in \bigcap_n V_n$. Then $F_n(x_0) > \frac{1}{2} I$ thus $F_n(x_0) \not\to 0$ and we obtain a contradiction.
