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I am trying to answer the following question: Let $g$ be a bounded lebesgue measurable function on $\mathbb{R}$ such that $\lim_{n} \int_I g(nx) dx = 0$ for all intervals $I \subset [0,1]$. Suppose $f \in L^1([0,1])$. Show $\lim_{n} \int_0^1 f(x) g(nx) dx = 0$.

I have the idea that I can treat $g(nx)$ as a radon nikodym derivative, use the jordan decomposition to split $g(nx)dx$ into two nonnegative measures, and show that those assign $0$ mass to the unit interval as $n$ goes to infinity. In this way the desired integral would also tend to $0$. But, I have not been able to properly flesh out the details and would like some help.

The reason I want to use the jordan decomposition to split the measure is so that I can use theorems requiring nonnegative functions like Fatou's lemma to show convergence. I am aware that there are proofs of this problem using other methods but I specifically want to use the idea above for practicing usage of radon nikodym.

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  • $\begingroup$ Presumably we have a restriction on $f$? Possibly $f \in L^1[0,1]$? $\endgroup$ Commented Aug 10, 2019 at 19:40
  • $\begingroup$ @BrianMoehring Yes, my bad. The problem says exactly that. I will edit the question. $\endgroup$ Commented Aug 10, 2019 at 19:42
  • $\begingroup$ What is the role of $n$ in $g(nx)$? $\endgroup$ Commented Aug 10, 2019 at 20:10
  • $\begingroup$ The limits are all taken over $n$. I was also told it would be good to think about this problem if you take $g$ to be something like $sin(n\pi x)$ to get some intuition. $\endgroup$ Commented Aug 10, 2019 at 20:22

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Since $\lim_{n\to\infty} \int_I g(nx)\, dx = 0$ for all intervals $I \subset [0,1]$,we get $$\lim_{n\to\infty}\int_0^1f(x)g(nx)\,dx=0$$ for any step function $f(x)=\sum_{k=1}^n a_k\chi_{I_k}$ where $I_k\subset[0,1]$ is an interval.

Now suppose $f\in L^1$. Since step functions are dense in $L^1$, for any $\epsilon>0$ we can write $f=f_1+f_2$ where $f_1$ is a step function and $\int_0^1|f_2|\,dx<\epsilon$. Using what we've proven for step functions and the boundedness of $g$ gives that $$\lim_{n\to\infty}\int_0^1f(x)g(nx)\,dx=0.$$

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  • $\begingroup$ I'm trying to understand where radon nikodym comes in. When you say $L^1$ do you mean $L^1([0,1], \mu)$, where maybe $\mu$ is the measure $g(nx)dx$ converges to (since space of borel measures is weak star compact this limit exists). $\endgroup$ Commented Aug 11, 2019 at 0:18
  • $\begingroup$ @user9781778 When I say $L^1$ I mean $L^1([0,1],\lambda)$ where $\lambda$ is the Lebesgue measure. Btw, I can't see what has this problem to do with Radon-Nikodym. $\endgroup$ Commented Aug 11, 2019 at 0:30
  • $\begingroup$ I'm just saying since $g(nx)$ is a bounded measurable function for each $n$, we can think of them as radon nikodym derivatives. And my assessment is that if we integrate $f$ against those measures, we should get $0$ as $n$ increases since I think the limit assumption we're given can be thought of as saying that the measures eventually pick up no mass on the unit interval. $\endgroup$ Commented Aug 11, 2019 at 0:35
  • $\begingroup$ @user9781778 I see. In my opinion, you can understand this problem in this way. But to prove it, it may not be an efficient thought. I've proved your problem, using the Lebesgue theory. Hopefully you can understand it. $\endgroup$ Commented Aug 11, 2019 at 0:43
  • $\begingroup$ Oh I totally get what you have written; thanks. I think maybe the first two lines can be used to prove it the way I am thinking about if I can properly justify passing the limit through the integral. $\endgroup$ Commented Aug 11, 2019 at 0:45

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