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Problem $1$ from UCI's Fall 2015 Real Analysis Qualifying Exam asks us to calculate $$\lim_{n\to\infty}\int_E f(nx)dx$$ given that $f\in C(\mathbb R)$ is $1$-periodic and $E\subseteq [0,2\pi]$ is measurable.

According to the solutions website, the limit is $\mu(E)\int_0^1 f(x)dx$, where (I think) they meant to write the Lebesgue measure $m$ instead of $\mu$. Unfortunately, I don't see how one could guess that this is the correct limit, let alone prove it. The solution provided in the website defines $$h(x)=f(x)-\int_0^1 f(t)dt$$ and notes that $h$ is also $1$-periodic. It then writes $$\int_E f(nx)dx = \int_E\left(h(nx)+\int_0^1 f(t)dt\right)dx = \int_E h(nx)dx +\mu(E)\int_0^1 f(t)dt$$ and notes that it suffices to prove that $\lim_{n\to\infty}\int_E h(nx)dx=0$. The proof seems to make sense, but I have two questions:

  1. How could one guess that the limit is $\mu(E)\int_0^1 f(x)dx$? The limit $\lim_{n\to\infty}\int_E f(nx)dx$ reminded me of a similar limit $\lim_{n\to\infty} \int_0^1 f(x)\sin(nx)dx$ occurring in James Stewart's Calculus: Early Transcendentals, which I correctly computed to be $0$. Given that the graph of $f(nx)$ is "squishing together" as $n\to\infty$, I thought the limit $\lim_{n\to\infty}\int_E f(nx)dx$ would also be $0$. This was until I noticed I could add $1$ to $f(nx)$ to get $1+f(nx)$, which is still $1$-periodic but gives $$\lim_{n\to\infty}\int_E (1+f(nx))dx=\mu(E)+\int_E f(nx)dx$$ which is more easily believed to be nonzero.

  2. Where does $h$ come from? I'm not sure what thought process could lead someone to come up with $h$ and see its significance for the solution. Is there an intuitive way of thinking about what $h$ is doing here?

Any and all responses are greatly appreciated.

Alann Rosas
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    If you think in Fourier modes, every mode is made to oscillate infinitely (and so converges weakly to $0$, the positive and negative parts cancel) except the mode of order $0$. – LL 3.14 Apr 24 '25 at 11:22
  • If you're looking for intuition, start with simpler cases. Instead of an arbitrary $E$, use an interval. Then draw a picture: as $n$ gets larger, what happens to the graph? What do you see about the periods of $f$ in the interval? – Brian Moehring Apr 24 '25 at 15:55
  • @LL3.14 when you say “Fourier modes”, are you referring to the terms in the Fourier series $f(x) = \sum_{k=-\infty}^\infty c_k e^{2\pi i k x}$? – Alann Rosas Apr 24 '25 at 16:00
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    The intuition is the same as what you see in the Riemann-Lebsgue Lemma. The rapid oscillations average out in the limit. It similar to the ergodic theorem in a way. – Mittens Apr 24 '25 at 16:18
  • @Mittens that’s what I was thinking after reading LL3.14’s comment. Is the idea to expand $f$ into a Fourier series, then integrate termwise and use the Riemann-Lebesgue lemma? – Alann Rosas Apr 24 '25 at 17:09
  • @BrianMoehring at least if $E$ is of the form $[0, a]$, it makes sense that the limit should be $\mu(E) \int_0^1 f(x)dx$. After substituting $u=nx$, we get $a\cdot \frac{1}{an}\int_0^{na} f(u)du$, or $\mu(E)\cdot \frac{1}{an}\int_0^{na} f(u)du$, and intuitively $\frac{1}{an}\int_0^{na} f(u)du$ tends to $f_{\text{ave}}([0,\infty))$. But since $f$ is periodic, the average on $[0,\infty)$ is the same as that on $[0,1]$. – Alann Rosas Apr 24 '25 at 17:20
  • Not quite. One has to do some calculations and see how the averaging happens. things in terms of probability. The rapid oscillations are like -1,1's in in Bernoulli samples. See the paper of Fejer Lebesguessche Konstanten und divergente Fourierreihen, Journal für die reine und angewandte Mathematik, vol. 1910, no. 138, 1910, pp. 22-53. – Mittens Apr 24 '25 at 17:23

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