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Let $ g(x) $ be a continuous periodic function of period 1 on $\mathbb{R}$. Prove that for any integrable function $f(x)$ on $[0,1]$,
$$ \lim_{n \to \infty}\int_0^{1}f(x)g(nx)dx= \int_0^{1}f(x)dx \int_0^{1}g(x)dx.$$

Any help is appreciated.

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3 Answers

Begin with a transformation $u=nx.$

$$\int_{0}^1 f(x)g(nx)dx = \frac{1}{n} \int_{0}^n f(u/n)g(u)du.$$

Break the integrand up into a sum of intervals.

$$ \frac{1}{n} \int_{0}^n f(u/n)g(u)du=\frac{1}{n}\sum_{j=1}^n\int_{j-1}^{j} f(u/n)g(u)du.$$

Make another variable transformation: $v=u-(j-1).$ Because $g(u)$ is 1 periodic $g(u)=g(u+1)=g(u+j-1).$

$$ \frac{1}{n}\sum_{j=1}^n\int_{j-1}^{j} f(u/n)g(u)du = \frac{1}{n}\sum_{j=1}^n\int_{0}^{1} f \left(\frac{u-(j-1)}{n} \right)g(u-(j-1))du $$

Rearrange the terms. $$ \frac{1}{n}\sum_{j=1}^n\int_{0}^{1} f\left(\frac{u-(j-1)}{n} \right)g(u-(j-1))du =\int_{0}^1 \left(\frac{1}{n} \sum_{j=1}^n f \left(\frac{u-(j-1)}{n} \right)\right)g(u)du .$$

We have now produced a Riemann sum which converges to an integral in the limit. We are now allowed, by dominated convergence theorem, to say

$$ \begin{align*} \lim_{n\to \infty} \int_{0}^1 f(x)g(nx)dx &= \int_{0}^1 \left(\int_{0}^1 f(z)dz\right) g(u)du \\ &=\int_{0}^1 f(z)dz\int_{0}^1 g(u)du=\int_{0}^1 f(x)dx\int_{0}^1 g(x)dx . \end{align*}$$

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Thank you so much. –  Sume Aug 10 '11 at 3:28
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this is not a good proof, just an idea. First consider f to be simple function, we can calculate that this identity holds. Then by the definition of Lebesgue integration, we can approximate f by simple functions. Because g is continuous(bounded), so there's no problem for the left hand side to converge.

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Ok. I have seen this problem in the book: Principles of Real Analysis by C.D.Aliprantis and O.Burkinshaw, and since I knew that this book has a solution manual, I went and searched over there and got the solution. The problem given in the book is as follows:

$\textbf{Problem.}$ Let $f:(0,\infty) \to \mathbb{R}$ be a real-valued continuous function such that $f(x+1)=f(x)$ for all $x \geq 0$. If $g:[0,1] \to \mathbb{R}$ is an arbitrary continuous function, then show that $$\lim_{n \to \infty} \ \int\limits_{0}^{1} g(x) \cdot f(nx) \ dx = \Biggl( \ \ \int\limits_{0}^{1} g(x) \ dx \Biggr) \cdot \Biggl( \ \ \int\limits_{0}^{1} f(x) \ dx\Biggr)$$

$\textbf{Solution.}$ Please see the book: Problems in real analysis a workbook with solutions Problem 23.14 Page $\textbf{205}$ for a complete solution.

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