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Let $f \in L^1(\mathbb{R})$. Find $$ \lim_{n \rightarrow \infty} \int_{-\infty}^\infty f(x)\sin(nx) dx \,. $$

Here's an attempt:

If we use a simple function approximation, we end up with something of the form $$\int_\mathbb{R} \sum_{i=1}^{k}c_i \chi_{E_i}(x)\sin(nx)dx = \sum_{i=1}^k c_i \int_{E_i} \sin(nx)dx$$ which goes to $0$ as $n \to \infty$ (The $E_i$'s can be chosen to be disjoint). My issue is that at least one of these $E_i$'s would necessarily be infinite. How do I fix this?

Does it then suffice to take a sequence of simple functions $s_m$ such that $$||s_m - f ||_{L^1} < \epsilon$$ for $n \geq N$ and each $s_m$ agrees with $s_{m-1}$ in it's sum except for the infinite term in which it splits the infinite $E_i$ further? That is, suppose $$s_{m-1} = \sum_{i=1}^{k} c_i \chi_{E_i}$$ with $||s_{m-1} - f||_{L^1} < \epsilon$. Wlog, suppose $E_k$ has infinite Lebesgue measure. Then surely there exists $s_m$ with $$ s_m = s_{m-1} + c_k'\chi_{E_k'} + c_{k+1}\chi_{E_{k+1}}$$ where $E_k' \cup E_{k+1} = E_k$ and $||s_m - f||_{L^1} < \epsilon$. Does this suffice? Can I then proceed to claim that

$$\lim_{n \to \infty} \int_\mathbb{R} (\lim_{m \to \infty} s_m) \sin(nx) \, dx = 0$$ by swapping the sum and the limits to get $$\sum_{i=1}^{\infty} \lim_{n \to \infty}\int_{E_i} c_i \sin (nx) \, dx$$ with $\mu (E_i) < \infty$? Or is this swapping not justified?

The previous post about this question is here.
Summary of prev post: Use the uniform continuity of the linear functional $$\mathcal{F}(f) = \lim_{n \rightarrow \infty} \int_{-\infty}^\infty f(x)\sin(nx) dx$$ to extend to all of $L^1$ using the density of something.

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  • $\begingroup$ Why would you want an infinite $E_i$? That would give you a function not in $L^1$. $\endgroup$ – Robert Israel Oct 23 '15 at 5:15
  • $\begingroup$ @RobertIsrael Poor wording. By "need" I meant that if I split $\mathbb{R}$ into finitely many disjoint intervals, one $E_i$ would necessarily be infinite. I want to avoid this. $\endgroup$ – Anthony Peter Oct 23 '15 at 5:19
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Hint: there is a finite interval $E$ such that $\int_{E^c} |f(x)| \; dx < \epsilon$. Then approximate the restriction of $f$ to $E$ by simple functions...

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  • $\begingroup$ Is it not enough to say that since $s \in L^1$, either $\mu(E_i) < \infty$ forall $i$, or $c_i = 0$ for $\mu(E_i) = \infty$? Or is this essentially saying the same thing since $f$ can be approximated by $s$? I think I'm trying to say this is reverse and less elegantly... I believe what I've said in this comment implies that such an $E$ exists implicitly $\endgroup$ – Anthony Peter Oct 23 '15 at 6:09
  • $\begingroup$ Namely, if $||f-s||_{L^1} < \epsilon$, then if $$E = \bigcup_{i=1}^{k} E_i$$ $$\int_{E} |f| dx < \epsilon.$$ $\endgroup$ – Anthony Peter Oct 23 '15 at 6:15
  • $\begingroup$ This is part of how you approximate an $L^1$ function by a simple function. $\endgroup$ – Robert Israel Oct 23 '15 at 6:17
  • $\begingroup$ So then I believe on this $E$ the integral is easily found to be $0$ and we can extend properly to all of $L^1$. Is there anything else that I need take into consideration? $\endgroup$ – Anthony Peter Oct 23 '15 at 6:19
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I'd first prove the limit statement for $f\in L^1$ that is also a step function: $f(x) = \sum_{k=1}^n 1_{(a_k,b_k]}(x)\cdot c_k$, where $a_1<b_1\le a_2<b_2<\cdots\le a_n<b_n$, and the $c_k$ are real. Then show that a general $f\in L^1$ can be approximated in $L^1$ by such step functions. For this second step it might be helpful to first approximate $f\in L^1$ by continuous functions of compact support.

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