Can someone provide a proof of the Riemann Lebesgue Lemma that doesn't make use of the known integral of $e^{itx}$? In particular I'm looking for the solution to the following problem:
Show that $\hat{f}(t)$ (the Fourier transform) “vanishes a infinity” on R. Hint: the Lebesgue measure is “outer regular”. We also have that the Fourier transform is a bounded continuous function, that $f(x) \in \mathcal{L}^1(\mathbb{R},B)$, and that $f(x)e^{itx}$ is also integrable.
Outer regularity means that:
$\forall A$ $\mu(A) = inf\{\mu(B): B $ is open and contains $A \}$.
Intuitively this happens because the wavelength of each wave becomes arbitrarily small, and I think the way I'm suppose to approach is is to replace $f(x)$ with a function that approximates it, so we have the integral of some simple function $f_n$ times $e^{itx}$, but I don't know how to use outer regularity, or even move in a useful direction from here.
(As a note, this was homework, but it has already been turned in. Just really want to know how to do it!)
For compactly supported smooth functions, you can truly integrate by parts. Doing so and taking absolute values, you get a factor of $1/|\xi|$ times an integral involving the derivative which is convergent.
Then you have that compactly supported smooth functions are dense in $L^1$. Picking an $\varepsilon$-approximation $g$ of your given $L^1$ function $f$, you can do the usual triangle inequality trick:
$$\left | \hat{f}(\xi) \right | = \left | \int f e^{-i \xi x} \right | = \left | \int (f-g) e^{-i \xi x} + \int g e^{-i \xi x} \right | \leq \left | \int (f-g) e^{-i \xi x} \right | + \left | \int g e^{-i \xi x} \right |.$$
Now the second term is already tending to zero by the integration by parts, while the first term is (regardless of $\xi$) at most $\varepsilon$ by the $L^1$ approximation and a triangle inequality use.
(This is pretty much what the Wikipedia article says, by the way.)