# limit of the integral of function evaluated at $x^n$

This is a problem from Problems in Mathematical Analysis: Integration by Kaczor and Nowak:

For a function $f$ continuous on $[0,1]$, find $$\lim_{n\to\infty}\int_0^1f(x^n)dx.$$

Here is the given solution:

Let $0<\epsilon<1$. Then $$\int_0^1f(x^n)dx=\int_0^{1-\epsilon}f(x^n)dx+\int_{1-\epsilon}^1f(x^n)dx$$and, by the first mean value theorem, $$\int_0^{1-\epsilon}f(x^n)dx=f(\xi^n)(1-\epsilon,\quad\text{where }0\le\xi\le(1-\epsilon).$$Thus $$\lim_{n\to\infty}f(0)(1-\epsilon).$$Moreover, $$\left|\int_{1-\epsilon}^1f(x^n)dx\right|\le M\epsilon,\quad\text{where }M=\sup\lbrace|f(x)|:x\in[0,1]\rbrace.$$Consequently, $$\lim_{n\to\infty}\int_0^1f(x^n)dx=f(0).$$

I understand everything except the last part, how do we conclude that $\lim_{n\to\infty}\int_0^1f(x^n)dx=f(0)$?

Anyway, I think we can use the same argument to show that $\lim_{n\to\infty}\int_{1-\epsilon}^1f(x^n)dx=f(0)\epsilon$. If this is correct, why don't we just choose $\epsilon=\frac12$ from the start, to make it simpler?

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I think that this is because $0 \leq x < 1$ on $[0,1)$ and the above argument allows us to interchange the limit and the integral. – Aleks Vlasev Nov 11 '11 at 4:02
"$\lim\limits_{n\to\infty}f(0)(1-\epsilon)$" is not a statement. The sentence beginning "Thus" is rather a fragment. – Jonas Meyer Nov 11 '11 at 4:04
Note that: $$\left|\lim_{\epsilon\rightarrow0}\int_{1-\epsilon}^1f(x^n)dx\right|=\lim_{\eps‌​ilon\rightarrow0}\left|\int_{1-\epsilon}^1f(x^n)dx\right|\leq \lim_{\epsilon\rightarrow0} (M\epsilon) = 0 \quad\Rightarrow\quad \lim_{\epsilon\rightarrow0}\int_{1-\epsilon}^1f(x^n)dx=0$$ – matt Nov 11 '11 at 4:30

Since $\xi \in [0,1)$, then $\xi^n \to 0$. Hence, $f(\xi^n) \to f(0)$ as $f$ is continuous.
An alternative proof: Since $f$ is continuous on $[0,1]$, then $f$ is uniformly continuous on $[0,1]$. Since $f$ is uniformly continuous, we can switch the integral and limit:
\begin{align} \lim_{n \to \infty} \int_0^{1} f(x^n) dx &= \int_{0}^1 \lim_{n \to \infty} f(x^n)dx \\ &= \lim_{\epsilon \to 0} \int_0^{1 - \epsilon} \lim_{n \to \infty}f(x^n) dx \\ &= \lim_{\epsilon \to 0} \int_0^{1 - \epsilon} f(0) dx \\ &= \lim_{\epsilon \to 0}(1 - \epsilon) f(0) \\ &= f(0). \end{align}