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Consider $f:\mathbb{R}\to \mathbb{C}$ a bounded and $1$-periodic function, and $g \in L^1(R)$ then

$$\lim_{n\to \infty} \int _{R}g(x)f(nx)dx=\int_0^1f(s)ds\int_R g(t)dt.$$

I think the fact that $f$ is complex valued as well $g$ is not a big deal but I couldn't solve it in any case.

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1 Answer 1

By the density of simple functions it suffices to consider the case $g=\chi_{[a,b]}$. When $n$ is large, the integral of $f(nx)$ over $[a,b]$ splits into about $n(b-a)$ equal integrals over intervals of length $1/n$, plus one small interval which does not contribute much. So you get $(b-a)\int_0^1 f(x)dx$ in the limit.

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