To give background to my question, in all the books I've looked at to derive the inverse Fourier transform of a continuous function $f$ on $\mathbb{R}$, they seem to work as follows. Let $k$ be a positive real number and write (by standard Fourier series theory)
$$
f(x) = \sum_{n=-\infty}^\infty c_n \, e^{i n \pi x/k}\ ,
$$
where
$$
c_n = \frac{1}{2k} \int_{-k}^k f(y) \, e^{-i n \pi y/k}\, dy\
$$
Let $\xi_n = n \pi / k$ and $\Delta \xi_n = \xi_n - \xi_{n-1} = \pi/k$, then
$$
f(x) = \frac{1}{2\pi} \sum_{n=-\infty}^\infty
\left( \int_{-k}^k f(y) \, e^{-i y \xi_n}\, d y \right) e^{i x \xi_n } \Delta \xi_n.
$$
Take $k \to \infty$ to get
$$
f(x) = \frac{1}{2 \pi} \int_{-\infty}^\infty \widehat{f}(\xi)\, e^{i x \xi}\, d \xi
$$
where
$$
\widehat{f}(\xi) = \int_{-\infty}^\infty f(x)\, e^{-ix \xi} \,d x.
$$
Everyone does this argument, but they never prove it. Even if a book says this argument is not rigorous, they never once give a counterexample.
My Question is about the proof: Let $g(\xi)$ be absolutely integrable on $\mathbb{R}$ and continuous, but NOT compactly supported, for otherwise my question is trivial. Let $k > 0$, let $\xi_n = n/k$, and consider the Riemann sum $$ \sum_{n = -\infty}^\infty g(\xi_n)\Delta \xi_n $$ where suppose that this infinite series is absolutely convergent for all $k > 0$. Must it be true that $$ \lim_{k \to \infty} \sum_{-\infty}^\infty g(\xi_n)\Delta \xi_n = \int_{-\infty}^\infty g(\xi) d \xi? $$ If this is true, can you give an elementary proof, a proof using only the theory of the Riemann integral and no measure theory or other advanced stuff? If not, what is an explicit and not too complicated counterexample?
Certainly convergence of the Riemann sums does not follow from integrability and continuity of $g$.
Consider this function $g$:
Spikes of decreasing width. Area of the spike centered at $m\pi$ is $1/2^m$. We have:
$\bullet $ $g$ is continuous
$\bullet $ $\int_{-\infty}^{+\infty} |g(\xi)|\;d\xi = 1$
$\bullet $ $g(m\pi) = 1$ for $m=1,2,3,\dots$;
$\bullet $ $g(x) \ge 0$ for $-\infty < x < \infty$.
Now fix $k \in \mathbb N$. For $m=1,2,3\dots$ we have $\xi_{mk} = m\pi$, and there are infinitely many of these, so $$ \sum_{n=-\infty}^{+\infty} g(\xi_n)\Delta \xi_n = \frac{\pi}{k}\sum_{-\infty}^{+\infty} g(\xi_n) \ge \frac{\pi}{k}\sum_{m=1}^\infty g(m\pi) = +\infty . $$