Timeline for answer to Thinking of $\int f(x) g(nx) dx$ as an integral with respect to the measure $g(nx)dx$ by Feng

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Aug 11, 2019 at 0:45	comment	added	user9781778		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.
Aug 11, 2019 at 0:43	comment	added	Feng		@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.
Aug 11, 2019 at 0:35	comment	added	user9781778		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.
Aug 11, 2019 at 0:30	comment	added	Feng		@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.
Aug 11, 2019 at 0:18	comment	added	user9781778		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).
Aug 10, 2019 at 23:35	history	answered	Feng	CC BY-SA 4.0