Now, I know that the question was answered here: $\int_0^\infty ne^{-nx}\sin\left(\frac1{x}\right)\;dx\to ?$ as $n\to\infty$
But I'm looking for more of a measure theoretical approach, so if someone can point me in the right direction, I would be grateful. (i.e. use of Dominated Convergence Theorem or Monotone Convergence Theorem)
My rough idea has been the following:
We can't use MCT because the functions are not pointwise increasing. We can't use DCT since there there is no function to bound $ne^{-nx} sin(1/x)$.
But, it seems that on $[\ln 2, \infty)$, $e^{-x} \geq ne^{-nx}$, and on $[\ln (\frac{n+1}{n}, \ln \frac{n}{n-1}]$, $ne^{-nx} \geq me^{-mx}$ for all $n, m \in \mathbb{N}$ (I don't have proof of this, and it seems hard to prove).
So we can use DCT on integral restricted such intervals. That is, we can use DCT on the integral $$\int_{\mathbb{R}} ne^{-nx} \sin(\frac{1}{x}) \chi_{[\ln(n+1/n), \ln (n/n-1)]}dx$$ And on these intervals, the integrals are all 0.
Since these intervals partitions $[0, \infty)$, it follows our integral is 0.
I don't know if the idea is correct, and even if it was, it wouldn't be an elegant solution. So I'm hoping someone could point me in the right direction for a clean solution.
Thanks!
EDIT:
If we let $u = nx$, we get the integral
$$\int^\infty_0 e^{-u} \sin(\frac{n}{u}) du$$.
We could apply DCT to this to get 0?
Another EDIT:
But $\sin(\frac{n}{u})$ is not convergent, so we cannot apply DCT.