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?

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    $\begingroup$ $f$ is continuous and defined on a compact set hence has a maximum $M$. Note that $g:[0,1] \to \mathbf R$ defined by $g(x) = M$ for all $x \in [0,1]$ is a dominating function for $f_n$ (that is $f_n(x) \leqslant g(x)$ for all $x \in [0,1])$. Hence we can apply Lebesgue Dominated Convergence and use continuity and that $0 \leqslant x < 1$ to obtain the result. That $f(1^n) = f(1)$ for all $n$ doesn't matter since $\{1\}$ is a set of measure zero in $[0, 1]$. $\endgroup$ – Jonas Teuwen Oct 10 '11 at 22:21
  • $\begingroup$ I'm in the Riemann Theory. $\endgroup$ – Mec Oct 10 '11 at 22:32
  • $\begingroup$ Doesn't matter, if $f$ is Riemann integrable on $[0, 1]$ it also is Lebesgue integrable. If you want a more elementary argument check out the post by Phira. $\endgroup$ – Jonas Teuwen Oct 10 '11 at 22:45
  • $\begingroup$ yeah, I'm not allowed to use that theory $\endgroup$ – Mec Oct 10 '11 at 22:49

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 $n$.

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    $\begingroup$ To elaborate: let $0 < \epsilon < 1$ be small. Let $\delta > 0$ then we have for large $n$ that $f(0) - \delta < f(x^n) < f(0) + \delta$ for all $0 \leqslant x < 1$ (why?). Now bound the integral from $0$ to $1 - \epsilon$. For the integral from $1 - \epsilon$ to $\epsilon$ let $M$ be the maximum of $f$ and then... $\endgroup$ – Jonas Teuwen Oct 10 '11 at 22:40
  • $\begingroup$ actually that was one of my strategies too, but I got stucked in proving that for large $n$ we have $f(0)-\delta<f(x^n)<f(0)+\delta$ for all $0\leq x<1$. Could you help me? $\endgroup$ – Mec Oct 10 '11 at 22:53
  • $\begingroup$ For large n and x smaller than 1- epsilon, $x^n$ is small, for small argument, the function is near $f(0)$. $\endgroup$ – Phira Oct 10 '11 at 22:58
  • $\begingroup$ so you are saying that $f_n(x)$ is converging to $f(0)$ uniformly on $[0,1-\varepsilon]$? $\endgroup$ – Mec Oct 12 '11 at 23:21
  • $\begingroup$ @Alex Yes, because the speed of convergence is bounded below by the speed at 1- epsilon. $\endgroup$ – Phira Oct 13 '11 at 5:58

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.


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