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$\textbf{The Problem:}$ Prove that if $f$ is continuous on $[0,1]$, then $$\lim_{n\to\infty}\int_0^1f(x^n)\,dx=f(0).$$ Note: Here we are using the Riemann integral.

$\textbf{My Thoughts and Attempt:}$ Choose $0<\delta<1$. Since $f$ is continuous on a compact interval, it follows that it is also uniformly continuous, on $[0,1]$ and hence, on $[0,\delta]$. Thus, $f(x^n)\to f(0)$ uniformly on $[0,\delta]$. Therefore, $$\lim_{n\to\infty}\int_0^\delta f(x^n)\,dx=f(0)\cdot\delta.$$ Since $0<\delta<1$ was arbitrary, it follows that $$\lim_{n\to\infty}\int_0^1f(x^n)\,dx=f(0).$$

$\textbf{My Concerns:}$ I am not entirely convinced that I have executed my $\delta$ trick, which turns the problem into a double limit, correctly. Could anyone please provide some feedback on my proof above and whether the introduction of $\delta$ is handled appropriately?

Thank you for your time and appreciate any feedback.

Stackman
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1 Answers1

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The lines after "as $0<\delta<1$ was arbitrary is fuzzy. You do not need a double limit to conclude.

The idea is a good one though. I would split the integral into two pieces, one on which you can argue by uniform continuity, and one on which you can make $\delta$ small enough so that it does not matter. This I believe is what you are missing.

Fix $1>\delta>0$. Examine $$ \lim_{n\to \infty}\int_0^1f(x^n)\mathrm dx= \lim_{n\to \infty}\int_0^{1-\delta}f(x^n)\mathrm dx+ \lim_{n\to \infty}\int_{1-\delta}^1 f(x^n)\mathrm dx $$ Provided both exist. On the first, argue as you did. On the second, bound the integral uniformly in $n$ using the maximum $M=\max_{x\in [0,1]}|f(x)|$. Then, $$ \sup_n \left| \int_{1-\delta}^1f(x^n)\right|\leq M\delta $$

Now you can say since your choice of $\delta$ was arbitrary, the limit is what it should be.

operatorerror
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