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For a continuous $f(x)$ on $[0,1]$ I need to calculate $\lim_{n\to \infty}\int^1_0 f(x^n)dx\,$

Let $\epsilon>0$ then,$\quad \int^1_0 f(x^n)dx\,=\int^{1-\epsilon}_0 f(x^n)dx\, +\int^1_{1-\epsilon}f(x^n)dx\,$, Applying mean value theorem $\exists c_1, c_2 $ such that $\int^{1-\epsilon}_0 f(x^n)dx\,=f(c_1^n)(1-\epsilon)$ and $\int^1_{1-\epsilon}f(x^n)dx\,=f(c_2^n)\epsilon$ , then $\lim_{n\to \infty}\int^1_0 f(x^n)dx\,=\lim_{n\to \infty} (f(c_1^n)(1-\epsilon)+f(c_2^n)\epsilon)=f(0)$

Did
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Mathronaut
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1 Answers1

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I don't know if this is the Riemann integral, but here is a way I was thinking about doing this using Lebesgue Dominated Convergence Theorem. I want to define my sequence of continuous functions to be,

$$ f_n(x)=f(x^n) $$

We see that on the interval $[0,1]$ this sequence of functions converges to $f(0)$ almost everywhere because for almost all $x\in [0,1]$,

$$\lim_{n\longrightarrow \infty}x^n=0 \quad \text{pointwise}$$ and continuity of $f$ allows us to move the limit by,

$$\lim_{n\longrightarrow \infty}f(x^n)=f(\lim_{n \longrightarrow \infty} x^n)=f(0)$$

for almost all $x\in [0,1]$.

Since $f$ is continuous on $[0,1]$ we know that it achieves a maximum and minimum value on the interval. Take,

$$M=\max_{x\in [0,1]}|f(x)| $$

Then we have for each $x\in [0,1]$,and for each natural number $n$ that $|f_n(x)|=|f(x^n)|\leq M$ and clearly $M\in L^1([0,1])$ since,

$$\int_{0}^1Mdx=M < \infty $$

So we can apply LDCT to move the limit inside the integral sign, and by the continuity of $f$ we may therefore move it inside $f$ to get,

\begin{align*}\lim_{n\longrightarrow \infty}\int_0^1f_n(x)dx&=\\ \lim_{n\longrightarrow \infty}\int_0^1f(x^n)dx&=\\ \int_0^1 \lim_{n\longrightarrow \infty}f(x^n)dx&=\\ \int_0^1 f(\lim_{n\longrightarrow \infty}x^n)dx&=\\ \text{(almost everywhere)}\quad\int_0^1f(0)dx&=\\ f(0) \end{align*}

monroej
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